Calculate the amount of methane formed by the liberation of $149.6 \ kJ$ of heat using the following equation:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ/mol$ (in $g$)

The bond dissociation enthalpies of $H_2, Cl_2$ and $HCl$ are $434, 242$ and $431 \, kJ/mol$ respectively. The enthalpy of formation of $HCl$ is ..... $kJ/mol$.

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$
Assertion $(A)$: Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always $-57 \ kJ \ mol^{-1}$.
Reason $(R)$: Enthalpy of neutralisation is the amount of heat liberated when one mole of $H^{+}$ ions furnished by acid combine with one mole of $OH^{-}$ ions furnished by base to form one mole of water.
In the light of the above statements,choose the correct answer from the options given below.

If the bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio of $1 : 1 : 0.5$ and $\Delta_f H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$,the bond dissociation energy of $X_2$ will be in $kJ \ mol^{-1}$:

On the basis of the following thermochemical data: $(\Delta_fG^o H^{+}_{(aq)} = 0)$
$H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \,; \, \Delta H = 57.32 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} \,; \, \Delta H = -286.20 \, kJ$
The value of enthalpy of formation of $OH^{-}$ ion at $25 \, ^oC$ is : .............. $kJ$

Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$(I)$ $C_{\text{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \ kJ$
$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$
$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H = -1410.8 \ kJ$ (in $kJ$)