Physical Chemistry - GRE

Card 0 of 196

Question

Which of the following sets of quantum numbers () could correspond to a orbital?

Answer

Since we're looking at the orbital, we know . The range of possible values for is 0 to . Possible values for range to . Therefore, among the answer choices, is the only possible combination of quantum numbers corresponding to an orbital.

Compare your answer with the correct one above

Question

Which of the following sets of quantum numbers could correspond to a orbital?

Answer

Since we're looking at the orbital, we know . The range of possible values for is 0 to . Possible values for range to . Therefore, among the answer choices,

is the only possible combination of quantum numbers corresponding to a orbital.

Compare your answer with the correct one above

Question

For any given chemical reaction, one can draw an energy diagram. Energy diagrams depict the energy levels of the different steps in a reaction, while also indicating the net change in energy and giving clues to relative reaction rate.

Below, a reaction diagram is shown for a reaction that a scientist is studying in a lab. A student began the reaction the evening before, but the scientist is unsure as to the type of the reaction. He cannot find the student’s notes, except for the reaction diagram below.

Upon further review, the scientist realizes that the reaction in question involved formation of a carbocation that quickly reacted again to form stable products. At which point would we most likely find this carbocation in the above diagram?

Answer

Point 3 is where you would expect to find a relatively stable intermediate. An intermediate is more stable than a transition state, but not as stable as the original reactants and final products. Stability is inversely proportional to energy, thus we are looking for the point that is between the highest and lowest energies in the reaction. By this logic, point 1 is the reactants, 2 and 4 are transition states, 3 is a stable intermediate, and 5 is the products.

Compare your answer with the correct one above

Question

What kinetic equation describes the rate of an E1 process?

Answer

Most organic reactions are carried out in multiple steps. The rate equation can be derived from the process that occurs in the slowest step of the mechanism. An E1 reaction's slow step is when a leaving group separates from the hydrocarbon and a carbocation is formed. The only reactant in this step is one molar equivalent of the hydrocarbon. Thus, the rate equation only depends on that substance. The molar equivalent determines the exponent of each reactant in the equation.

Compare your answer with the correct one above

Question

$A + 2B + C \rightarrow D$

You want to find the order of each reactant by manipulating the reactant concentrations in multiple trials. The table below shows the effect of altered reactant concentrations on initial reaction rate.

$\begin{matrix} \text{Trial} & \text{Initial [A]} & \text{Initial [B]} & \text{Initial [C]} & \text{Rate}\ (\frac{M}{s})\ 1&0.1 &0.3 &0.2 &0.002 \ 2& 0.1& 0.9& 0.2& 0.006\ 3& 0.2& 0.3& 0.2& 0.002\ 4& 0.1& 0.3& 0.4 & 0.008 \end{matrix}$

Using the above trials, write the rate law for the reaction.

Answer

Keep in mind that the chemical equation and its coefficients have NOTHING to do with the rate law. The order of each reactant must be determined by experiment.

To find the order of each reactant, compare the initial reaction rates of two trials in which only one of the three reactants' concentrations is altered. For example, trials 1 and 4 keep A and B equal, but C is doubled. When C is doubled, we see that the initial reaction rate is quadrupled. As a result, we determine that the order of reactant C is 2. When the reactant is altered, but the initial reaction rate is kept constant, as seen in trials 1 and 3 with respect to A, the order of that reactant is 0. Finally, when the reactant is multiplied by the same factor that the initial reaction rate is multiplied, as seen in trials 1 and 2 with respect to B, the order of the reactant is 1.

Putting the data together: A is zeroth order, B is first order, and C is second order. Our rate law can thus be written .

Compare your answer with the correct one above

Question

A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:

$NH_{4}^{+}(aq) + NO_{2}^{-}(aq) \rightarrow N_{2}(g) +2H_{2}O(l)$

In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.

$\begin{matrix} Trial & [NH_4^+] & [NO_2^-] &Rate \ 1& 0.480M &0.120M &0.018\frac{M}{s} \ 2& 0.240M & 0.120M& 0.009\frac{M}{s}\ 3& 0.240M& 0.360M & 0.027\frac{M}{s} \end{matrix}$

Reaction rates depend on reaction concentrations, as well as the reaction constant . The Arrhenius equation is used to calculate this constant. Which of the following factors is used in the Arrhenius equation?

I. Temperature

II. Activation energy

III. Reactant concentrations

Answer

If you thought all of these factors are used in the Arrhenius equation, consider that the rate constant is independent of the reactant concentrations. The concentration effect is considered in the rate law by multiplying the concentrations by the constant, but the concentrations themselves are independent of the actual Arrhenius calculation.

The calculation shows that $k=Ae^{-\frac{E_a}{RT}}$ . In this equation, $\small E_a$ is the activation energy and $\small T$ is the temperature.

Compare your answer with the correct one above

Question

A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:

$NH_{4}^{+}(aq) + NO_{2}^{-}(aq) \rightarrow N_{2}(g) +2H_{2}O(l)$

In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.

$\begin{matrix} Trial & [NH_4^+] & [NO_2^-] &Rate \ 1& 0.480M &0.120M &0.018\frac{M}{s} \ 2& 0.240M & 0.120M& 0.009\frac{M}{s}\ 3& 0.240M& 0.360M & 0.027\frac{M}{s} \end{matrix}$

Which of the following most closely approximates the rate law for this reaction?

Answer

The reaction table in the passage indicates that the reaction rate varies in a 1-to-1 fashion as you vary the each reactant, while holding the other constant.

The rate law is written as .

Compare trials 1 and 2 to see that doubling the ammonium concentration doubles the rate. The reaction is first order for ammonium: .

Compare trials 2 and 3 to see that tripling nitrate concentration triples the rate. The reaction is first order for nitrate: .

The final rate law is .

Compare your answer with the correct one above

Question

For the reaction, $2X\ +\ Y\rightarrow\ X_{2}Y$ , the initial rate of the reaction was determined and the values are tabulated below.

What is the overall order for the rate law of this reaction?

Answer

For the reaction given:

$rate\ of\ reaction=k[X]^{a}[Y]^{b}$

We need to find the reaction order, therefore we have to find the values of a and b in the equation. First, let us compare the results of the experiments done to determine how changing the concentration effects the rate of the reaction. Let's compare run number 1 with run number 2 in which the concentration of Y was doubled in run 2 while keeping the concentration of X constant. We can observe that the reaction rate was quadrupled. Let's compare by ratios:

$\frac{rate_{2}}{rate_{1}}=\left ( \frac{[X]_{2}^{a}[Y]_{2}^{b}}{[X]_{1}^{a}[Y]_{1}^{b}} \right )$

The value for the concentration of B was the same for run 1 and 2 and therefore they cancel out:

$\frac{rate_{2}}{rate_{1}}=\left ( \frac{[Y]_{2}^{b}}{[Y]_{1}^{b}} \right )$

Plugging the values into the equation gives:

$\frac{3.0\frac{M}{s}}{0.75\frac{M}{s}}=\frac{(0.01M)^{b}}{(0.005M)^{b}}$

Simplifying the above equation gives:

$4.0=2.0^{b}$

Therefore,

We can also compare the results of run number 2 and run number 3. In that case, the concentration of Y was kept constant while the concentration of X was doubled. We can see that the reaction rate doubled. Therefore the reaction is first order with respect to X and second order with respect to Y. Therefore, the reaction is third order overall:

$rate\ of\ reaction=k[X][Y]^{2}$

Compare your answer with the correct one above

Question

For the reaction, $A\ +\ B\rightarrow\ C\ +\ D$ , the initial rate of the reaction was determined and the values are tabulated below.

Determine the overall order for the rate law of this reaction.

Answer

For the reaction given:

$rate\ of\ reaction=k[A]^{x}[B]^{y}$

We need to find the reaction order, therefore we have to find the values of x and y in the equation. First, let us compare the results of the experiments done to determine how changing the concentration effects the rate of the reaction. Let's compare run number 1 with run number 2 in which the concentration of A was doubled in run 2 while keeping the concentration of B constant. We can observe that the reaction rate was doubled. Let's compare by ratios:

$\frac{rate_{2}}{rate_{1}}=\left ( \frac{[A]_{2}^{x}[B]_{2}^{y}}{[A]_{1}^{x}[B]_{1}^{y}} \right )$

The value for the concentration of B was the same for run 1 and 2 and therefore they cancel out:

$\frac{rate_{2}}{rate_{1}}=\left ( \frac{[A]_{2}^{x}}{[A]_{1}^{x}} \right )$

Plugging the values into the equation gives:

$\frac{5.0\times 10^{-5}}{2.0\times 10^{-5}}=\frac{(4.0\times 10^{-3})^{x}}{(2.0\times 10^{-3})^{x}}$

Simplifying the above equation gives:

$2.0=2.0^{x}$

Therefore,

We can also compare the results of run number 2 and run number 3. In that case, the concentration of A was kept constant while the concentration of B was decreased by half. We can see that the reaction rate decreased by half. Therefore the reaction is first order with respect to A and first order with respect to B. Therefore, the reaction is second order overall:

Compare your answer with the correct one above

Question

$H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2} O\ \ \Delta H_{f}^{0} = - 286 kJ$

$C + O_{2}\rightarrow CO_{2} \ \ \Delta H_{f}^{0} = - 394 kJ$

$8C + 9H_{2}\rightarrow C_{8} H_{18}\ \ \Delta H_{f}^{0} = -269 kJ$

Given the enthalpies of formation, what is the enthalpy of combustion of octane in the reaction:

$2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2} +18H_{2}O$

Answer

The equation for enthalpy of reaction is:

$\Delta H_{reaction}^{0} = \Delta H_{f, products}^{0} - \Delta H_{f, reactants}^{0}$

Given our chemical reaction and the enthalpies of formation, we can find the enthalpy of reaction.

$2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2} +18H_{2}O$

First, find the total enthalpy for the products.

$\Delta H_{f, products}^{0} = (16 ) \Delta H_{f, CO_{2}} + (18 ) \Delta H_{f, H_{2}O}$

$\Delta H_{f, products}^{0} = (16 ) (-394kJ) + (18 )(-286kJ) = -11452 kJ$

Then, find the total enthalpy for the reactants.

$\Delta H_{f, reactants}^{0} = (2) \Delta H_{f, C_{8}H_{18}} + (25 ) \Delta H_{f, O_{2}}$

$\Delta H_{f, reactants}^{0} =( 2) (269kJ) + (25) (0kJ) = 538kJ$

Since the oxygen is elemental, its heat of formation is zero.

Return to the original equation to calculate the final enthalpy of reaction.

$\Delta H_{reaction}^{0}=-11452kJ-(538kJ)=-11990kJ\approx-1.2*10^{4}kJ$

Compare your answer with the correct one above

Question

Propane gas combusts according the following chemical equation:

$C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$

Given the following standard enthalpy values, what is the enthalpy of the reaction for the combustion of one mole of propane?

$\Delta H_{C_{3}H_{8}} = -103.8\frac{kJ}{mol}$

$\Delta H_{O_{2}} = 0\frac{kJ}{mol}$

$\Delta H_{CO_{2}} = -393.5\frac{kJ}{mol}$

$\Delta H_{H_{2}O} = -241.8\frac{kJ}{mol}$

Answer

Given the molar enthalpy values for reactants and products, we can solve for the enthalpy of the reaction using the equation:

$\Delta H_{reaction} = \Delta H_{products} - \Delta H_{reactants}$

Keep in mind that the molar enthalpy values must be multiplied by the coefficients that are present in the balanced reaction:

$C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$

$\Delta H_{reaction} = (3\Delta H_{CO_{2}} + 4\Delta H_{H_{2}O} ) - (\Delta H_{C_{3}H_{8}})$

Oxygen is omitted because its enthalpy value is zero.

$\Delta H_{reaction} = (3(-393.5\frac{kJ}{mol})+ 4(-241.8\frac{kJ}{mol})) - (-103.8\frac{kJ}{mol})$

$\Delta H_{reaction} = -2044kJ$

Compare your answer with the correct one above

Question

Which is not characteristic of an endothermic reaction?

Answer

An endothermic reaction absorbs energy in the form of heat. An endothermic reaction involves breaking of chemical bonds because it involves the absorption of energy. Endothermic reactions tend to feel cold because it is taking heat away from your skin. Since endothermic reactions involve absorbing energy, often in the form of heat, the change in enthalpy is positive.

Therefore, the answer is "releasing of energy to surroundings" which is not a characteristic of an endothermic process. Instead, it is a characteristic of an exothermic reaction which is the direct opposite of an endothermic reaction.

Compare your answer with the correct one above

Question

Which of the following reactions has a positive value for change in entropy $(\Delta S > 0)$ ?

Answer

Entropy is a measure of the disorder of a system. In other words, when the components of a system become more uniform and spread out, the entropy of the system has increased. For example, when a liquid becomes gaseous, the molecules separate from one another, increasing the disorder of the system. As a result, the evaporation of water results in an increase in entropy.

All other options result in a more ordered system, which decreases entropy.

Compare your answer with the correct one above

Question

At $25^{\circ}C$ , a reaction has a Gibb's free energy change of . If the enthalpy change of the reaction is , what is the entropy change of the reaction?

Answer

We can relate enthalpy, entropy, and Gibb's free energy using the Gibb's free energy equation:

$\Delta G = \Delta H-T\Delta S$

Keep in mind that temperature must be in Kelvin. Since we know enthalpy and Gibb's free energy for the reaction, we can solve for the entropy change:

$45,000J = 35,000J - (298K)(\Delta S)$

$\Delta S = -33.6\frac{J}{K}$

Compare your answer with the correct one above

Question

Which of the following options indicates that a chemical reaction is unfavorable?

Answer

The sign of $\Delta G$ indicates the direction of a chemical reaction and determines if a reaction is spontaneous or not. A chemical reaction in which $\Delta G< 0$ is considered to be a spontaneous reaction because it will proceed without requiring any outside energy. To approach this problem, we must consider the equation for the Gibbs free energy of a reaction, $\Delta G= \Delta H -T \Delta S$ . A reaction is considered to be favorable when $\Delta G$ is negative. The second law of thermodynamics indicates that the entropy of the universe is spontaneously increasing. This corresponds to a positive $\Delta S$ .

Compare your answer with the correct one above

Question

Use the following values for water as needed.

$\Delta H^o_{fus}=334\frac{J}{g}$

$\Delta H^o_{vap}=2260\frac{J}{g}$

$c=4.186\frac{J}{g^oC}$

$\rho = 1\frac{g}{mL}$

If burning wood releases $\small 20kJ$ of heat energy per gram of wood consumed, what mass of wood must be consumed to heat $\small 1L$ of water from $\small 20^oC$ to $\small 100^oC$ , and then to convert it to water vapor?

Answer

There are two processes requiring added heat in this problem:

1. Raising the temperature of the liquid water from $\small 20^oC$ to $\small 100^oC$ (use $\Delta Q = mc\Delta T$ )

2. Boiling the water at a constant temperature of $\small 100^oC$ (use $\Delta Q = m\Delta H^o_{vap}$ )

To use either of these equation, we need to find the mass of the water using the relation between mass, density, and volume.

$m = \rhoV= 1\frac{g}{mL}1000mL=1000g$

Use this mass with the given specific heat and temperatures to find the heat for part 1 of the process.

$\Delta Q = (1000g)(4.186\frac{J}{g^oC})(100^oC - 20^oC)=334,880J$

Then, use the mass with the given heat of vaporization to find the energy needed to convert the water to water vapor.

$\Delta Q =(1000g) (2260\frac{J}{g})= 2,260,000J$

Sum the energies for step 1 and step 2.

$334,880 J + 2,260,000 J = 2,594,880J\approx 2,595kJ$

This is the total amount of energy needed from the burning wood. Use stoichiometry to find the grams of wood needed to produce this amount of energy.

$2595kJ*(\frac{1 g}{20kJ})=129.7g\approx 130g$

Compare your answer with the correct one above

Question

Calculate the specific heat of of metal that requires of heat energy to raise its temperature from $20.46^{o}C$ to $64.4^{o}C$ ?

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

$C=\frac{Q}{m\bigtriangleup T}=\frac{410.4J}{(20.4g\times (64.4^{o}C-25.4^{o}C))}=0.516\frac{J}{g^{o}C}$

Compare your answer with the correct one above

Question

What was the final temperature of water if a sample of water absorbs of heat energy and heats up from $25^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$810.5J=19g\times4.179\frac{J}{g^{o}C}\times(T_{f}-25^{o}C)$

$810.5J=79.40\frac{J}{^{o}C}\times(T_{f}-25^{o}C)$

$\frac{810.5J}{79.40\frac{J}{^{o}C}}=T_{f}-25^{o}C$

$10.21^{o}C=T_{f}-25^{o}C$

$T_{f}=35^{o}C$

Compare your answer with the correct one above

Question

An metal at $54.2^{o}C$ was put into of water at $25^{o}C$ . The final temp of the water and metal was $29.3^{o}C$ . Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ .

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(81.9g\cdot C\cdot (29.3^{o}C-54.2^oC)=40.7g\cdot4.18\frac{J}{g^o}C(29.3^{o}C-25^{o}C)$

$-(-2039g^{o}C\cdot C)=731J$

$2039g^{o}C\cdot\ C=731J$

$C=\frac{731J}{2039g^{o}C}=0.358\frac{J}{g^{o}C}$

Compare your answer with the correct one above

Question

Calculate the final temperature when a sample of metal (specific heat of the metal= $0.52 \frac{J}{g^{o}C}$ ) at $45.0^{o}C$ is placed into of water at $23.0^{o}C$ . (specific heat of water is $4.18\frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(44.0g\cdot 0.52\frac{J}{g^{o}C}\cdot (T_{f}-45.0^oC)=95.0g\cdot4.18\frac{J}{g^o}C(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}(T_{f}-45.0^{o}C))=397\frac{J}{^{o}C}(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-22.9\frac{J}{^{o}C}\cdot 45.0^{o}C)=397\frac{J}{^{o}C}\cdot T_{f}-397\frac{J}{^{o}C}\cdot 23.0^{o}C$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-1031J)=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}+1031J=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}-397\frac{J}{^{o}C}\cdot T_{f}=-1031J-9131J$

$-419\frac{J}{^{o}C}\cdot T_{f}=-10162J$

$T_{f}=\frac{-10162^{o}C}{-419.9}=24^{o}C$

Compare your answer with the correct one above

Tap the card to reveal the answer