Given, $\Delta \mathrm{H}=170 \times 10^{3} \mathrm{J}$
$\Delta S=170 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$
Applying, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S},$ the value of $\Delta \mathrm{G}=$ -ve only when $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H},$ which is possible only when $\mathrm{T}=1110 \mathrm{K}$
$\Delta G=170 \times 10^{3}-(1110 \times 170)$$=-18700 \mathrm{J}$
Thus, reaction is spontaneous at $\mathrm{T}=1110 \mathrm{K}$
$(i)$ $H_{(aq)}^+ + OH^-= H_2O_{(l)} ,$ $\Delta H = -X_1\,kJ \,mol^{-1}$
$(ii)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} = H_2O_{(l)},$ $\Delta H = -X_2\,kJ \,mol^{-1}$
$(iii)$ $CO_{2(g)} + H_{2(g)} = CO_{(g)} + H_2O_{(l)},$ $\Delta H = -X_3\, kJ\, mol^{-1}$
$(iv)$ $ C_2H_{2(g)}+ \frac{5}{2} O_{2(g)} = 2CO_{2(g)} + H_2O_{(l)},$ $\Delta H = -X_4\,kJ \,mol^{-1}$
તો $H_2O_{(l)}$ સર્જનઉષ્મા કેટલી હશે ?