6.2. Equilibrium - II (icon Equilibrium) — Chemistry STD 11 Science — Question

$\begin{array}{l} Cl - Cl ( g ) \longrightarrow Cl ^*( g )+ Cl ^*( g ) \Delta H ^{\circ}=58 kcal mol ^{-1} \\ H _3 C - Cl ( g ) \longrightarrow H _3 C ^*( g )+ Cl ^{\circ}( g ) \Delta H ^{\circ}=85 kcal mol ^{-1} \\ H - Cl ( g ) \quad \longrightarrow H ^*( g ) \quad+ Cl ^*( g ) \Delta H ^{\circ}=103 kcal mol ^{-1} \\\end{array}$

($1$) Correct match of the $C - H$ bonds (shown in bold) in Column $J$ with their BDE in Column $K$ is

$(A)$ $P - iii, Q - iv, R - ii, S - i$

$(B)$ $P - i, Q - ii, R - iii, S - iv$

$(C)$ $P - iii, Q - ii, R - i, S - iv$

$(D)$ $P - ii, Q - i, R - iv, S - iii$

($2$) For the following reaction

$CH _4( g )+ Cl _2( g ) \xrightarrow{\text { light }} CH _3 Cl ( g )+ HCl ( g )$

the correct statement is

$(A)$ Initiation step is exothermic with $\Delta H ^{\circ}=-58 kcal mol ^{-1}$

$(B)$ Propagation step involving ${ }^{\circ} CH _3$ formation is exothermic with $\Delta H ^{\circ}=-2 kcal mol ^{-1}$.

$(C)$ Propagation step involving $CH _3 Cl$ formation is endothermic with $\Delta H ^{\circ}=+27 kcal mol ^{-1}$.

$(D)$ The reaction is exothermic with $\Delta H ^{\circ}=-25 kcal mol ^{-1}$.