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MCAT Physical : Enthalpy

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Example Question #1 : Enthalpy

The formation of nitrous oxide is a 2-step process.

$N_{2} + O_{2} \rightarrow 2NO (step 1)$ $\displaystyle N_{2} + O_{2} \rightarrow 2NO (step 1)$

$2NO + O_{2} \rightarrow 2NO_{2} (step 2)$ $\displaystyle 2NO + O_{2} \rightarrow 2NO_{2} (step 2)$

$N_{2} + 2O_{2} \rightarrow 2NO_{2} (overall reaction)$ $\displaystyle N_{2} + 2O_{2} \rightarrow 2NO_{2} (overall reaction)$

The overall enthalpy of the reaction is +68kJ.

If the enthalpy change of step 1 is 180kJ, what is the enthalpy change of step 2?

Possible Answers:

248kJ

–112kJ

112kJ

More information is needed to answer the question.

Correct answer:

–112kJ

Explanation:

Hess's law states that the enthalpies of the steps in an overall reaction can be added in order to find the overall enthalpy. Since the overall enthalpy of the reaction is +68kJ, we can use the following equation to find step 2's enthalpy.

$Step 1 \Delta H + step 2 \Delta H = overall reaction \Delta H$ $\displaystyle Step 1 \Delta H + step 2 \Delta H = overall reaction \Delta H$

$180kJ + step 2 \Delta H = +68kJ$ $\displaystyle 180kJ + step 2 \Delta H = +68kJ$

$Step 2 \Delta H = -112kJ.$ $\displaystyle Step 2 \Delta H = -112kJ.$

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Example Question #2 : Enthalpy

Consider the following processes:

1) $A \rightarrow 2B + C + \Delta H_{1}$ $\displaystyle A \rightarrow 2B + C + \Delta H_{1}$

2) $D\rightarrow B + C + \Delta H_{2}$ $\displaystyle D\rightarrow B + C + \Delta H_{2}$

3) $D \rightarrow 2E + \Delta H_{3}$ $\displaystyle D \rightarrow 2E + \Delta H_{3}$

Which for the following expresses $\Delta H$ $\displaystyle \Delta H$ for the process $A + C \rightarrow 4E$ $\displaystyle A + C \rightarrow 4E$ ?

Possible Answers:

$\Delta H_{1} + \Delta H_{2} - 2\Delta H_{3}$ $\displaystyle \Delta H_{1} + \Delta H_{2} - 2\Delta H_{3}$

$\Delta H_{1} - 2\Delta H_{2} + 2\Delta H_{3}$ $\displaystyle \Delta H_{1} - 2\Delta H_{2} + 2\Delta H_{3}$

$\Delta H_{1} + 2\Delta H_{2} + 2\Delta H_{3}$ $\displaystyle \Delta H_{1} + 2\Delta H_{2} + 2\Delta H_{3}$

$\Delta H_{1} + 2\Delta H_{2} - 2\Delta H_{3}$ $\displaystyle \Delta H_{1} + 2\Delta H_{2} - 2\Delta H_{3}$

$\Delta H_{1} + 2\Delta H_{3}$ $\displaystyle \Delta H_{1} + 2\Delta H_{3}$

Correct answer:

$\Delta H_{1} - 2\Delta H_{2} + 2\Delta H_{3}$ $\displaystyle \Delta H_{1} - 2\Delta H_{2} + 2\Delta H_{3}$

Explanation:

1) $A \rightarrow 2B + C + \Delta H_{1}$ $\displaystyle A \rightarrow 2B + C + \Delta H_{1}$

2) $D\rightarrow B + C + \Delta H_{2}$ $\displaystyle D\rightarrow B + C + \Delta H_{2}$

3) $D \rightarrow 2E + \Delta H_{3}$ $\displaystyle D \rightarrow 2E + \Delta H_{3}$

$A + C \rightarrow 4E$ $\displaystyle A + C \rightarrow 4E$

We need to add the given processes in such a way that leaves $\small A+C$ $\displaystyle \small A+C$ on the left side, and $\small 4E$ $\displaystyle \small 4E$ on the right.

Since there are no $\small B's$ $\displaystyle \small B's$ in the final equation, begin by combining processes 1 and 2 to eliminate.

First, reverse process 2 and multiply by two:

2a) $2 (B + C + \Delta H_{2} \rightarrow D)$ $\displaystyle 2 (B + C + \Delta H_{2} \rightarrow D)$

2a) $2B + 2C + 2\Delta H_{2} \rightarrow 2 D$ $\displaystyle 2B + 2C + 2\Delta H_{2} \rightarrow 2 D$

2a) $2B + 2C \rightarrow 2 D - 2\Delta H_{2}$ $\displaystyle 2B + 2C \rightarrow 2 D - 2\Delta H_{2}$

Add this result to process 1.

1) $A \rightarrow 2B + C +\Delta H_{1}$ $\displaystyle A \rightarrow 2B + C +\Delta H_{1}$

2a) $2B + 2C \rightarrow 2D -2\Delta H_{2}$ $\displaystyle 2B + 2C \rightarrow 2D -2\Delta H_{2}$

1+2a) $A + C \rightarrow 2D + \Delta H_{1} -2\Delta H_{2}$ $\displaystyle A + C \rightarrow 2D + \Delta H_{1} -2\Delta H_{2}$

Now we need to combine with process 3 to eliminate the $\small D's$ $\displaystyle \small D's$. Start by multiplying process 3 by two.

3a) $2(D \rightarrow 2E + \Delta H_{3})$ $\displaystyle 2(D \rightarrow 2E + \Delta H_{3})$

3a) $2D \rightarrow 4E + 2\Delta H_{3}$ $\displaystyle 2D \rightarrow 4E + 2\Delta H_{3}$

Add this to the combined result from processes 1 and 2.

1+2a) $A + C \rightarrow 2D + \Delta H_{1} -2\Delta H_{2}$ $\displaystyle A + C \rightarrow 2D + \Delta H_{1} -2\Delta H_{2}$

3a) $2D \rightarrow 4E + 2\Delta H_{3}$ $\displaystyle 2D \rightarrow 4E + 2\Delta H_{3}$

1+2a+3a) $A + C \rightarrow 4E +\Delta H_{1} - 2\Delta H_{2} + 2\Delta H_{3}$ $\displaystyle A + C \rightarrow 4E +\Delta H_{1} - 2\Delta H_{2} + 2\Delta H_{3}$

This gives our final combination. The elemental reaction is $A + C \rightarrow 4E$ $\displaystyle A + C \rightarrow 4E$, and the enthalpy of the process is $\small \Delta H_1-2\Delta H_2+2\Delta H_3$ $\displaystyle \small \Delta H_1-2\Delta H_2+2\Delta H_3$.

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MCAT Physical : Enthalpy

Study concepts, example questions & explanations for MCAT Physical

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