Question

the table shows the enthalpy values for the different bonds found in the reaction ch₄ + 2 o₂ → co₂ + 2 h₂o. 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=(2×799)+ δh=(4×799)+ δh=(4×413)+ δh=(2×413)+

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. The enthalpy change $\Delta H$ of a reaction is given by $\Delta H=\text{Bonds broken}-\text{Bonds formed}$. Bonds broken: 4 C - H and 2 O=O, Bonds formed: 2 C=O and 4 O - H. $\Delta H=(4\times413 + 2\times495)-(2\times799+4\times463)$. But considering the general form for bond - enthalpy calculations of $\Delta H=\sum(\text{Bonds broken})-\sum(\text{Bonds formed})$, we focus on the formula structure for bonds broken first. For $CH_4$, 4 C - H bonds are broken so contribution from $CH_4$ to bonds broken is $4\times413$.

Answer:

$\Delta H=(4\times413)+(2\times495)-(2\times799)-(4\times463)$ (but from the given options, the part related to bonds broken in $CH_4$ is $\Delta H=(4\times413)+\cdots$ so the relevant option from the given ones is the one starting with $(4\times413)$)