Question

enthalpy of reaction
reaction energy
the table shows the enthalpy values for the different bonds found in the reaction ch₄ + 2o₂ → co₂ + 2h₂o
bond enthalpy (kj/mol)
c - h 413
o = o 495
c = o 799
o - h 463
which equation would allow you to calculate the enthalpy of the reaction?
δh=(4×799)+(2

Explanation:

Step1: Identify bonds broken and formed

In $CH_4 + 2O_2
ightarrow CO_2+2H_2O$, in $CH_4$ there are 4 C - H bonds, in $2O_2$ there are 2 O=O bonds. In $CO_2$ there are 2 C=O bonds and in $2H_2O$ there are 4 O - H bonds.

Step2: Use bond - enthalpy formula

The enthalpy of reaction $\Delta H$ is given by $\Delta H=\sum(\text{Bond enthalpies of bonds broken})-\sum(\text{Bond enthalpies of bonds formed})$. Bonds broken: 4 C - H and 2 O=O. Bonds formed: 2 C=O and 4 O - H. So $\Delta H=(4\times413 + 2\times495)-(2\times799+4\times463)$.

Answer:

$\Delta H=(4\times413 + 2\times495)-(2\times799+4\times463)$