GRE Subject Test: Chemistry - Enthalpy

Which is not characteristic of an endothermic reaction?

Releasing of energy to surroundings

Absorption of energy from surroundings

Breaking of a chemical bond

A chemical reaction that feels cold

The net change of enthalpy is positive

Explanation

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.

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

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

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

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

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

$-1.2*10^{4}kJ$

$1.2*10^{4}kJ$

$-1.09*10^{4}kJ$

$1.09*10^{-4}kJ$

Explanation

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.

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

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?

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

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

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

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

Explanation

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.

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

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

Enthalpy

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