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

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(A) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.Let the given reactions be:$(

Vedclass · Accepted Answer

$-393.7$

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(A) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.Let the given reactions be:$(i) C(	ext{graphite}) + \frac{1}{2} O_{2(g)} 	o CO_{(g)}; \Delta H_1 = -110.5 \ kJ$$(ii) CO_{(g)} + \frac{1}{2} O_{2(g)} 	o CO_{2(g)}; \Delta H_2 = -283.2 \ kJ$Adding equation $(i)$ and $(ii)$:$C(	ext{graphite}) + \frac{1}{2} O_{2(g)} + CO_{(g)} + \frac{1}{2} O_{2(g)} 	o CO_{(g)} + CO_{2(g)}$$C(	ext{graphite}) + O_{2(g)} 	o CO_{2(g)}$The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 = -110.5 \ kJ + (-283.2 \ kJ) = -393.7 \ kJ$.

Given the thermochemical reactions:
$C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta H = -110.5 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta H = -283.2 \ kJ$
Calculate the heat of reaction for $C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}$ in $kJ$.

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Given the thermochemical reactions: $C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta H = -110.5 \ kJ$ $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta H = -283.2 \ kJ$ Calculate the heat of reaction for $C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}$ in $kJ$.