Question

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:

Answer 1

ANSWER

The heat of reaction of the above reaction is -234 KJ

EXPLANATION

Given information

$\begin{gathered} 4NO_{(g)}\text{ + 2O}_{2(g)}\text{ }\rightarrow4NO_{2(g)}\text{ ------------ }\Delta H\degree_{rxn}\text{ =?} \\ N_{2(g)\text{ }}+\text{ O}_{2(g)}\text{ }\rightarrow2NO_{(g)}\text{ -------------- }\Delta H\degree_{rxn}\text{ = +185 KJ} \\ \frac{1}{2}N_{2(g)}\text{ + O}_{2(g)}\text{ }\rightarrow NO_{2(g)\text{ ------------------- }}\Delta H\degree_{rxn\text{ = }}+33\text{ }KJ \end{gathered}$

Recall, that change in enthalpy of the reaction is the summation of the products - summation of the reactants

Mathematically,

$\Delta H_{rxn}\text{ = }\Sigma\text{ }\Delta H_{products}\text{ - }\Sigma\text{ }\Delta H_{reactants}$

The next step is to write the formula for calculating the enthalpy

$\Delta H_{rxn\text{ }}\text{ = 4}(NO_2)\text{ - 4}(NO)\text{ + 2}(O_2)$

The next step is to substitute the given data into the above formula

$\begin{gathered} \Delta H_{rxn}\text{ = 4}(33)\text{ - }\lbrack4(91.5)\text{ - 2}(0)\rbrack \\ \Delta H_{rxn}\text{ = 132 - 366} \\ \Delta Hrxn\text{ = -234 KJ} \end{gathered}$

Hence, the heat of reaction of the above reaction is -234 KJ