Heat of combustion $\Delta H^\circ$ for $C(s)$, $H_2(g)$ and $CH_4(g)$ are $-94$, $-68$ and $-213 ext{ Kcal/m

Question

Heat of combustion $\Delta H^\circ$ for $C(s)$, $H_2(g)$ and $CH_4(g)$ are $-94$, $-68$ and $-213	ext{ Kcal/mol}$. $\Delta H^\circ$ for $C(s) + 2H_2(g) ightarrow CH_4 (g)$ is:

Accepted Answer

1. **Identify the Goal:** Calculate the enthalpy change ($\Delta H_{reaction}$) for the formation of methane ($CH_4$) using heat of combustion data.
   Reaction: $C(s) + 2H_2(g) ightarrow CH_4(g)$
2. **Apply Formula:** When heat of combustion ($\Delta H_c$) data is given, the enthalpy of reaction is calculated as:
   $$ \Delta H_{reaction} = \sum \Delta H_c(	ext{Reactants}) - \sum \Delta H_c(	ext{Products}) $$
3. **Substitute Values:**
   - Reactants: 1 mole of $C(s)$ and 2 moles of $H_2(g)$.
   - Product: 1 mole of $CH_4(g)$.
   $$ \Delta H_{reaction} = [1 \cdot \Delta H_c(C) + 2 \cdot \Delta H_c(H_2)] - [1 \cdot \Delta H_c(CH_4)] $$
   $$ \Delta H_{reaction} = [(-94) + 2(-68)] - [-213] $$
4. **Calculate:**
   $$ \Delta H_{reaction} = [-94 - 136] + 213 $$
   $$ \Delta H_{reaction} = -230 + 213 = -17	ext{ Kcal} $$

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