# What is an example of an endothermic process practice problem?

Jan 2, 2015

LONG ANSWER. Here are some of the questions you could get in an endothermic process problem:

You are given the following chemical rection

${N}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to 2 N {O}_{\left(g\right)}$

Provide an explanaition for why this reaction is endothermic (both conceptual, and mathematical);
Is this reaction spontaneous at 298 K?
If not, at what temperature does it become spontaneous?

Data given: $\Delta {H}_{f}^{\circ} = + 90.4 \text{kJ/mol}$ for $N O$ and DeltaS_("reaction") = 24.7 "J/K"

Let's start with the math to get it out of the way. A reaction is said to be endothermic if its change in enthalpy, $\Delta {H}_{\text{reaction}}$, is positive. We can calculate this change in enthalpy from what the data provides us.

$\Delta {H}_{\text{reaction") = "2 moles NO" * 90.4 (kJ)/(mol) - ("1 mole" N_2) * 0 (kJ)/(mol) - ("1 mole} {O}_{2}} \cdot 0 \frac{k J}{m o l}$

DeltaH_("reaction") = "2 moles NO" * 90.4 (kJ)/(mol) = 180.8 "kJ

The trick here was to be aware of the fact that the enthalpy of formation ($\Delta {H}_{f}^{\circ}$) for elements is zero.

Since $\Delta {H}_{\text{reaction}} > 0$, the reaction is indeed endothermic.

Conceptually, this reaction is endothermic despite the fact that a bond (between $N$ and $O$) is formed; this happens because the ${N}_{2}$ molecule has its two atoms bonded together by a very strong triple bond, which means that more energy must be put into breaking this bond than is released when the $N O$ molecule is formed.

Now, in order for a reaction to be spontaneous, the sign of $\Delta G$ - the Gibbs free energy - must be negative at the given temperature.

We can therefore determine this reaction's spontaneity by using

$\Delta {G}_{\text{reaction") = DeltaH_("reaction") - T * DeltaS_("reaction}}$

$\Delta {G}_{\text{reaction}} = 180.8 \cdot {10}^{3} J - 298 K \cdot 24.71 \cdot \frac{J}{K} = 173.4 k J$

The reaction is not spontaneous at this temperature. We can determine at what temperature the reaction starts to be spontaneous by setting $\Delta {G}_{\text{reaction}} = 0$.

$0 = \Delta {H}_{\text{reaction") - T * DeltaS_("reaction}}$

$T = \left(\Delta {H}_{\text{reaction"))/(DeltaS_("reaction}}\right) = \frac{180.8 \cdot {10}^{3} J}{24.71 \frac{J}{K}} = 7317 K$

This reaction becomes spontaneous at $7317$ $\text{K}$.

As a conclusion, questions about endothermic or exothermic processes revolve around $\Delta H$, $\Delta S$, and $\Delta G$ - if a reaction's spontaneity is in question.