$HCl_{ ( g )}$ are $-105,-242,-394$ and $-92\,kJ\,mol ^{-1}$
respectively. The magnitude of enthalpy of the reaction given below is $......kJ\,mol ^{-1}$
(nearest integer) $CCl _{4( g )}+2 H _2 O_{( g )} \rightarrow CO _{2( g )}+4 HCl_{( g )}$
$=(-394+4 \times-92)-(-105+(2 \times-242))$
$=-173\,kJ / mol$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(i)$ $H_{(aq)}^+ + OH^-= H_2O_{(l)} ,$ $\Delta H = -X_1\,kJ \,mol^{-1}$
$(ii)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} = H_2O_{(l)},$ $\Delta H = -X_2\,kJ \,mol^{-1}$
$(iii)$ $CO_{2(g)} + H_{2(g)} = CO_{(g)} + H_2O_{(l)},$ $\Delta H = -X_3\, kJ\, mol^{-1}$
$(iv)$ $ C_2H_{2(g)}+ \frac{5}{2} O_{2(g)} = 2CO_{2(g)} + H_2O_{(l)},$ $\Delta H = -X_4\,kJ \,mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is
