MCAT Physical - Enthalpy

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

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

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

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

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?

–112kJ

112kJ

248kJ

More information is needed to answer the question.

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.

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

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

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

Consider the following processes:

$A \rightarrow 2B + C + \Delta H_{1}$
$D\rightarrow B + C + \Delta H_{2}$
$D \rightarrow 2E + \Delta H_{3}$

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

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

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

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

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

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

Explanation

$A \rightarrow 2B + C + \Delta H_{1}$
$D\rightarrow B + C + \Delta H_{2}$
$D \rightarrow 2E + \Delta H_{3}$

$A + C \rightarrow 4E$

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

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

First, reverse process 2 and multiply by two:

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

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

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

Add this result to process 1.

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

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

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

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

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

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

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

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

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

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

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

Enthalpy

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