AP Chemistry - Thermodynamics

Which of the following is a law of thermodynamics?

ΔH (system) + ΔH (surroundings) = ΔH (universe)

ΔS (system) + ΔS (surroundings) = ΔS (universe)

ΔE (surroundings) = ΔE (system)

The entropy of the universe is always decreasing

Explanation

The second law of thermodynamics states that the entropy of the system and the entropy of the surroundings is equal to the entropy of the universe. The rest of the answer choices are not one of the fundamental laws of themodynamics.

Which of the following is a law of thermodynamics?

ΔH (system) + ΔH (surroundings) = ΔH (universe)

ΔS (system) + ΔS (surroundings) = ΔS (universe)

ΔE (surroundings) = ΔE (system)

The entropy of the universe is always decreasing

Explanation

1. $\small Zn \rightarrow Zn^{2+} + 2e^{-}$ $\small \small E^{o} = 0.762 V$

2. $\small Cu \rightarrow Cu^{2+} + 2e^{-}$ $\small \small E^{o}= -0.340 V$

3. $\small Zn^{2+} + Cu \rightarrow Zn + Cu^{2+}$

What is $\small \small E_{cell}^{o}$ ? Is reaction 3 spontaneous?

$\small -1.1 V$ , nonspontaneous

$\small -1.1 V$ , spontaneous

$\small 1.1 V$ , nonspontaneous

$\small 1.1 V$ , spontaneous

Explanation

$\small E^{o}_{cell} = E_{red}^{o} + E_{ox}^{o}$

$\small E_{cell}^{o} = -0.762 V - 0.340 V$

$\small E^{o}_{cell} = -1.1 V$

There are two concepts to consider in this problem. First, the question asks for the $\small E^{o}_{cell}$ . Reaction 3 has $\small Zn^{2+}$ being reduced so the potential for the half reaction becomes negative. $\small Cu$ half reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

For the following cell reaction:

2 Al (s) + 3 Mn2+ (aq) -> 2 Al3+ (aq) + 3 Mn (s)

predict if the cell potential Ecell will be larger, or smaller than Eocell for the following conditions: \[Al3+\] = 1.0 M; \[Mn2+\] = 5.0 M.

Larger

Smaller

Identical

Can not be determined

Explanation

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

At these concentrations Q will become smaller, and thus log Q will become smaller. This will give rise to a larger cell potential.

Calculate the standard cell potential of the following reaction:

Cd(s) + MnO2 (s) + 4 H+ (aq) + -> Cd2+ (aq) + Mn2+ (aq) + 2 H2O (l)

Given:

MnO2 (s) + 4 H+ (aq) + 2e- -> Mn2+ (aq) + 2 H2O (l) Eo = 1.23 V

Cd2+ (aq) + 2 e- -> Cd (s) Eo = -0.40 V

1.63 V

0.83 V

-1.63 V

-0.83 V

0.0 V

Explanation

Eocell = Eo cathode - Eoanode

Eocell = 1.23 – (-0.40) = 1.63 V

A chemistry student is trying to calculate how long it will take a power source of to heat a sample of ice from $-35^{o}C$ to $75^{o}C$ . Given that the specific heat capacity of ice is $2.06\frac{J}{g^{o}C}$ , the specific heat capacity of liquid water is $4.184\frac{J}{g^{o}C}$ , and the heat of fusion for water is $334.16\frac{J}{g}$ , how long will this process take?

There is not enough information to determine the amount of time needed for the process described

Explanation

In order to solve this problem, we'll need to break it up into steps.

Step 1: Calculate the amount of energy necessary to raise the sample of ice from $-35^{o}C$ to $0^{o}C$ . To do this, we'll need to use the following equation:

$q_{1}=mc_{H_{2}O(s)}\Delta T$

$q_{1}=( 20\cdot 10^{3}g)\left ( 2.06\frac{J}{g^{0}C} \right )\left ( 35^{o}C \right )=1.44\cdot 10^{6}J$

Step 2: Calculate the amount of energy necessary to convert the sample at $0^{o}C$ from ice to water. We'll need to make use of the following equation:

$q_{2}=m\Delta H_{fus(H_{2}O)}$

$q_{2}=\left ( 20\cdot 10^{3}g \right )\left ( 334.16\frac{J}{g} \right )=6.68\cdot 10^{6}J$

Step 3: Calculate the amount of energy necessary to convert the sample of water from $0^{o}C$ to $75^{o}C$ .

$q_{3}=mc_{H_{2}O(l)}\Delta T$

$q_{3}=\left ( 20\cdot 10^{3}g \right )\left ( 4.184\frac{J}{g^{o}C} \right )\left ( 75^{o}C \right )=6.28\cdot 10^{6}J$

Step 4: Sum the amount of energy from the previous 3 steps. This value is the total amount of energy for the entire process.

$q_{Total}=q_{1}+q_{2}+q_{3}=1.44\cdot 10^{7}J$

Step 5: Now that we know the total amount of energy needed for the process, we need to calculate the time based on the amount of power provided.

$P=\frac{q_{Total}}{t}$

$t=\frac{q_{Total}}{P}=\frac{\left ( 1.44\cdot 10^{7}J \right )}{(100\frac{J}{s}) }=1.44\cdot 10^{5}s$

$t=(1.44\cdot10^{5}s)\left ( \frac{1hr}{3600s} \right )=40hrs$

There is not enough information to determine the amount of time needed for the process described

Explanation

In order to solve this problem, we'll need to break it up into steps.

Step 1: Calculate the amount of energy necessary to raise the sample of ice from $-35^{o}C$ to $0^{o}C$ . To do this, we'll need to use the following equation:

$q_{1}=mc_{H_{2}O(s)}\Delta T$

$q_{1}=( 20\cdot 10^{3}g)\left ( 2.06\frac{J}{g^{0}C} \right )\left ( 35^{o}C \right )=1.44\cdot 10^{6}J$

Step 2: Calculate the amount of energy necessary to convert the sample at $0^{o}C$ from ice to water. We'll need to make use of the following equation:

$q_{2}=m\Delta H_{fus(H_{2}O)}$

$q_{2}=\left ( 20\cdot 10^{3}g \right )\left ( 334.16\frac{J}{g} \right )=6.68\cdot 10^{6}J$

Step 3: Calculate the amount of energy necessary to convert the sample of water from $0^{o}C$ to $75^{o}C$ .

$q_{3}=mc_{H_{2}O(l)}\Delta T$

$q_{3}=\left ( 20\cdot 10^{3}g \right )\left ( 4.184\frac{J}{g^{o}C} \right )\left ( 75^{o}C \right )=6.28\cdot 10^{6}J$

Step 4: Sum the amount of energy from the previous 3 steps. This value is the total amount of energy for the entire process.

$q_{Total}=q_{1}+q_{2}+q_{3}=1.44\cdot 10^{7}J$

Step 5: Now that we know the total amount of energy needed for the process, we need to calculate the time based on the amount of power provided.

$P=\frac{q_{Total}}{t}$

$t=\frac{q_{Total}}{P}=\frac{\left ( 1.44\cdot 10^{7}J \right )}{(100\frac{J}{s}) }=1.44\cdot 10^{5}s$

$t=(1.44\cdot10^{5}s)\left ( \frac{1hr}{3600s} \right )=40hrs$

The following galvanic cell is created:

$2Au^{3+} + 3Cu \rightarrow 3Cu^{2+} + 2Au$ $E^{\circ} = 1.16V$

Which of the following takes place at the cathode?

Gold ions receive electrons

Copper loses electrons

Gold loses electrons

Copper gains electrons

Explanation

Reduction always takes place at the cathode. This means that a substance is receiving electrons.

$2Au^{3+} + 3Cu \rightarrow 3Cu^{2+} + 2Au$

In the reaction, the gold ions are receiving electrons in order to create gold atoms. This takes place at the cathode.

Copper ions and gold are products, and will not be oxidized or reduced. Copper is reduced at the anode during this reaction.

How many grams of Cr can be obtained by the electrolysis of a Cr(NO3)3 if 10 amps are passed through the cell for 6 hours?

38.8 g

19.4 g

103 g

12.5 g

56.3 g

Explanation

$Cr^{3+} + 3 e- \rightarrow Cr(s)$

$mass = \frac{10 A , x, 6 hr , x , \frac{3600 s}{1 hr}}{ \frac{96485 C}{mol e-} } , x , \frac{1 mol , Cr}{3 mol e-} , x , \frac{51.996 g , Cr}{1 mol , Cr}= 38.8 g$

1. $\small Zn \rightarrow Zn^{2+} + 2e^{-}$ $\small \small E^{o} = 0.762 V$

2. $\small Cu \rightarrow Cu^{2+} + 2e^{-}$ $\small \small E^{o}= -0.340 V$

3. $\small Zn^{2+} + Cu \rightarrow Zn + Cu^{2+}$

What is $\small \small E_{cell}^{o}$ ? Is reaction 3 spontaneous?

$\small -1.1 V$ , nonspontaneous

$\small -1.1 V$ , spontaneous

$\small 1.1 V$ , nonspontaneous

$\small 1.1 V$ , spontaneous

Explanation

$\small E^{o}_{cell} = E_{red}^{o} + E_{ox}^{o}$

$\small E_{cell}^{o} = -0.762 V - 0.340 V$

$\small E^{o}_{cell} = -1.1 V$

Thermodynamics

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