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AP Chemistry : Enthalpy

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Example Question #11 : Enthalpy

Which of the following is a true statement regarding the energy involved in the formation of Methane, CH₄, from graphite C_(s)and H_2(g)?

Possible Answers:

Energy will be consumed, then released

None of the other answers

Energy will be released, then consumed

Energy will be released

Energy will be consumed

Correct answer:

Energy will be consumed, then released

Explanation:

This problem requires a knowledge of the energy involved when bonds are broken versus when bonds are formed. Energy is released when bonds are formed, and energy is consumed when bonds are broken. Here, methane is being formed from C and H_2.Since the carbon source, graphite, is monoatomic the carbon will be involved only in the formation of C–H bonds on it's way to forming methane. The H however exists as H₂ so bonds will need to be broken to allow for C–H bonds to form. So first energy is consumed to break the H–H bonds, it is then released when the C–H bonds form.

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Example Question #12 : Enthalpy

H2O ice melts to liquid H2O [Heat is added]

Which of the following is correct?

I. Δ H is positive

II. ΔS is positive

III. ΔH is negative

IV. ΔS is negative

Possible Answers:

I only

II only

I & II

I and IV

I and III

Correct answer:

I & II

Explanation:

When heat is added, enthalpy increases, ΔH refers to enthalpy

ΔS refers to entropy, which is disorder; disorder increases as it moves from solid to liquid to gas state.

As ice goes to liquid water, entropy increases

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Example Question #11 : Enthalpy

Which of the following equations is correct for finding change in enthalpy?

Possible Answers:

$\Delta H _{products} - \Delta H _{reactants}$

$\Delta H _{products} * \Delta H _{reactants}$

$\frac{\Delta H _{products}}{\Delta H _{reactants}}$

$\Delta H _{reactants}-\Delta H _{products}$

$\Delta H _{products} + \Delta H _{reactants}$

Correct answer:

$\Delta H _{products} - \Delta H _{reactants}$

Explanation:

The change in enthalpy of a reaction is determined by the amount of heat gained when the products bonds form, versus the amount of heat lost to break the bonds in the reactants. The correct formula for the enthalpy of reaction is the difference between the enthalpy of the products and that of the reactants:

$\Delta H _{products} - \Delta H _{reactants}$

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Example Question #14 : Enthalpy

$S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}$ $\Delta H = x$

$S_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow SO_{3(g)}$ $\Delta H = y$

Based on the given reactions, what is the enthalpy change in the following reaction?

$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$

Possible Answers:

2x - y

y-x

x-y

2x-2y

2y-2x

Correct answer:

2y-2x

Explanation:

The reactions given show the heats of formation of all the reactants and products, and we know the heat of formation of $O_{2}$ is zero because it is a pure element. Formation of the product, $SO_{3}$ gives us . For the reactants, $SO_{2}$ gives us plus for $O_{2}$ . Recall that the overall reaction's change in enthalpy is:

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

Thus for this overall reaction:

$\Delta H_{reaction} = 2y-2x$

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AP Chemistry : Enthalpy

Study concepts, example questions & explanations for AP Chemistry

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Example Question #11 : Enthalpy

Example Question #12 : Enthalpy

Example Question #11 : Enthalpy

Example Question #14 : Enthalpy

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