Question

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

Answer 1

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

Step-by-step 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} \\ (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