Given the thermochemical reactions: C(text{gr | vedclass

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(B) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.
Step $1$: $C(	ext{grap

Vedclass · Accepted Answer

$-393.7$

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(B) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.
Step $1$: $C(	ext{graphite}) + \frac{1}{2}O_2 ightarrow CO; \Delta H_1 = -110.5 \, kJ$
Step $2$: $CO + \frac{1}{2}O_2 ightarrow CO_2; \Delta H_2 = -283.2 \, kJ$
Adding the two equations:
$C(	ext{graphite}) + \frac{1}{2}O_2 + CO + \frac{1}{2}O_2 ightarrow CO + CO_2$
$C(	ext{graphite}) + O_2 ightarrow CO_2$
The total enthalpy change $\Delta H = \Delta H_1 + \Delta H_2$
$\Delta H = -110.5 \, kJ + (-283.2 \, kJ) = -393.7 \, kJ$.

Given the thermochemical reactions: $C(\text{graphite}) + \frac{1}{2}O_2 \rightarrow CO; \Delta H = -110.5 \, kJ$ and $CO + \frac{1}{2}O_2 \rightarrow CO_2; \Delta H = -283.2 \, kJ$. Calculate the heat of reaction for $C(\text{graphite}) + O_2 \rightarrow CO_2$ in $kJ$.

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