Enthalpy change for the reaction,$\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$,is called:

In the reaction $C + 2S \to CS_2 + \Delta H$,$\Delta H$ is the

Given the following thermochemical equations:
$(i) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} ; \Delta H = -68.39 \, kcal$
$(ii) \ K_{(s)} + H_2O_{(l)} + aq \rightarrow KOH_{(aq)} + \frac{1}{2}H_{2(g)} ; \Delta H = -48.0 \, kcal$
$(iii) \ KOH_{(s)} + aq \rightarrow KOH_{(aq)} ; \Delta H = -14.0 \, kcal$
Calculate the heat of formation of $KOH_{(s)}$.

The heat of neutralization of $HCl$ by $NaOH$ is $-57.3 \, kJ/mol$. If the heat of neutralization of $HCN$ by $NaOH$ is $-12.1 \, kJ/mol$,then the enthalpy of dissociation of $HCN$ is $...... \, kJ$.

Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH = -111 \ kJ$. If $N_2O_5(s)$ is formed instead of $N_2O_5(g)$, what will be the value of $\Delta_rH$ in $kJ$? (Given: $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$ for $N_2O_5$)

Given: $2Zn + O_2 \rightarrow 2ZnO, \Delta G^o = -616 \, J$; $2Zn + S_2 \rightarrow 2ZnS, \Delta G^o = -293 \, J$; $S_2 + 2O_2 \rightarrow 2SO_2, \Delta G^o = -408 \, J$. The value of $\Delta G^o$ for the reaction $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$ is ....... $J$.