The value of $\Delta_{f} H^{o}$ for $NH_3$ is $-91.8 \ kJ \ mol^{-1}$. Calculate the enthalpy change for the following reaction: $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$

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The formation reaction for $1 \ mole$ of $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \to NH_{3(g)}$; $\Delta_f H^o = -91.8 \ kJ \ mol^{-1}$.
For the formation of $2 \ moles$ of $NH_3$,the reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta_r H^o = 2 \times (-91.8 \ kJ \ mol^{-1}) = -183.6 \ kJ \ mol^{-1}$.
The given reaction is the reverse of the formation of $2 \ moles$ of $NH_3$.
Therefore,for the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the enthalpy change is: $\Delta_r H^o = -(-183.6 \ kJ \ mol^{-1}) = +183.6 \ kJ \ mol^{-1}$.

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Nitroglycerine $(MW = 227.1)$ detonates according to the following equation:
$2 C_{3}H_{5}(NO_{3})_{3(l)} \longrightarrow 3 N_{2(g)} + \frac{1}{2} O_{2(g)} + 6 CO_{2(g)} + 5 H_{2}O_{(g)}$
The standard molar enthalpies of formation,$\Delta H_{f}^{\circ}$ for all the compounds are given below:
$\Delta H_{f}^{\circ}[C_{3}H_{5}(NO_{3})_{3}] = -364 \, kJ/mol$
$\Delta H_{f}^{\circ}[CO_{2(g)}] = -393.5 \, kJ/mol$
$\Delta H_{f}^{\circ}[H_{2}O_{(g)}] = -241.8 \, kJ/mol$
$\Delta H_{f}^{\circ}[N_{2(g)}] = 0 \, kJ/mol$
$\Delta H_{f}^{\circ}[O_{2(g)}] = 0 \, kJ/mol$
The enthalpy change when $10 \, g$ of nitroglycerine is detonated is $..... \, kJ$.

Identify the law for the statement "Overall the enthalpy change for a reaction is equal to the sum of enthalpy changes of individual steps in the reaction".

What is the value of $\Delta H^{\circ}$ for the formation of ethanol from ethene gas and liquid water from the following data (in $kJ$)?
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