Given below are two statements:
Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.
The correct answer is:

Consider the following reactions:
$(i)$ $H_{(aq)}^{+} + OH^{-}_{(aq)} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_1 \ kJ \ mol^{-1}$
$(ii)$ $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$
$(iii)$ $CO_{2_{(g)}} + H_{2_{(g)}} \longrightarrow CO_{(g)} + H_2O_{(l)}$,$\Delta H = -X_3 \ kJ \ mol^{-1}$
$(iv)$ $C_2H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + H_2O_{(l)}$,$\Delta H = -X_4 \ kJ \ mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is

Given:
$C + 2S \to CS_2 ; \Delta H_f^o = +117.0 \, kJ \, mol^{-1} \dots (1)$
$C + O_2 \to CO_2 ; \Delta H_f^o = -393 \, kJ \, mol^{-1} \dots (2)$
$S + O_2 \to SO_2 ; \Delta H_f^o = -297 \, kJ \, mol^{-1} \dots (3)$
The heat of reaction for $CS_2 + 3O_2 \to CO_2 + 2SO_2$ is:
.....$kJ \, mol^{-1}$

Which of the following values of heat of formation indicates that the product is least stable in $kcal$?

The heats of hydrogenation for $3-$-methylbutene and $2-$-pentene are $-30\, kcal/mol$ and $-28\, kcal/mol$ respectively. The heats of combustion of $2-$-methylbutane and pentane are $-784\, kcal/mol$ and $-782\, kcal/mol$ respectively. All the values are given under standard conditions. Taking into account that combustion of both alkanes gives the same products,what is $\Delta H$ (in $kcal/mol$) for the following reaction under same conditions?
$3-$-methylbutene $\rightleftharpoons$ $2-$-pentene

The heat of combustion of carbon monoxide at constant volume and at $17 \ ^oC$ is $-283.3 \ kJ$. Calculate its heat of combustion at constant pressure in $kJ$. $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$ (in $.5$)