Calculate the standard heat of formation of carbon disulphide $(l)$,given that the standard heat of combustion of carbon $(s)$,sulphur $(s)$ and carbon disulphide $(l)$ are $-393.3, -293.72$ and $-1108.76 \text{ kJ mol}^{-1}$ respectively.

Which of the following reactions satisfies the following conditions?
$(a)$ Heat of combustion reaction
$(b)$ Heat of formation reaction
$(c)$ An exothermic reaction
$(d)$ Not a neutralization reaction

Given:
$(I) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}; \Delta H^o_{298\ K} = -285.9 \ kJ \ mol^{-1}$
$(II) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H^o_{298\ K} = -241.8 \ kJ \ mol^{-1}$
The molar enthalpy of vaporisation of water will be $kJ \ mol^{-1}$.

The bond dissociation enthalpy of $X_2$,$\Delta H_{\text{bond}}^{\circ}$,calculated from the given data is $...$ $kJ \ mol^{-1}$. (Nearest integer)
$M^{+}X^{-}_{(s)} \rightarrow M^{+}_{(g)} + X^{-}_{(g)} \quad \Delta H_{\text{lattice}}^{\circ} = 800 \ kJ \ mol^{-1}$
$M_{(s)} \rightarrow M_{(g)} \quad \Delta H_{\text{sub}}^{\circ} = 100 \ kJ \ mol^{-1}$
$M_{(g)} \rightarrow M^{+}_{(g)} + e^{-}_{(g)} \quad \Delta H_{i}^{\circ} = 500 \ kJ \ mol^{-1}$
$X_{(g)} + e^{-}_{(g)} \rightarrow X^{-}_{(g)} \quad \Delta H_{\text{eg}}^{\circ} = -300 \ kJ \ mol^{-1}$
$M_{(s)} + \frac{1}{2}X_{2(g)} \rightarrow M^{+}X^{-}_{(s)} \quad \Delta H_{f}^{\circ} = -400 \ kJ \ mol^{-1}$
[Given : $M^{+}X^{-}$ is a pure ionic compound and $X$ forms a diatomic molecule $X_2$ in gaseous state]

Two reactions are given below:
$2 Fe_{(s)} + \frac{3}{2} O_{2_{(g)}} \rightarrow Fe_2 O_{3_{(s)}}, \Delta H^{o} = -822 \ kJ/mol$
$C_{(s)} + \frac{1}{2} O_{2_{(g)}} \rightarrow CO_{(g)}, \Delta H^{o} = -110 \ kJ/mol$
Then,the enthalpy change for the following reaction is:
$3 C_{(s)} + Fe_2 O_{3_{(s)}} \rightarrow 2 Fe_{(s)} + 3 CO_{(g)}$

The oxidizing power of chlorine in an aqueous solution can be determined by the parameters given below. What is the total energy change in $kJ \ mol^{-1}$ for the conversion of $\frac{1}{2} Cl_{2(g)}$ to $Cl^-_{(aq)}$?
$\frac{1}{2} Cl_{2(g)}$ $\xrightarrow{\frac{1}{2} \Delta_{diss} H^\Theta} Cl_{(g)}$ $\xrightarrow{\Delta_{eg} H^\Theta} Cl^-_{(g)}$ $\xrightarrow{\Delta_{hyd} H^\Theta} Cl^-_{(aq)}$
(Given: $\Delta_{diss} H_{Cl_2}^\Theta = 240 \ kJ \ mol^{-1}$,$\Delta_{eg} H_{Cl}^\Theta = -349 \ kJ \ mol^{-1}$,$\Delta_{hyd} H_{Cl^-}^\Theta = -381 \ kJ \ mol^{-1}$)