Question
$${H_2}$$ gas is mixed with air at $${25^ \circ }C$$ under a pressure of 1 atmosphere and exploded in a closed vessel. The heat of the reaction, $${H_{2\left( g \right)}} + \frac{1}{2}{O_{2\left( g \right)}} \to {H_2}{O_{\left( \nu \right)}}$$ at constant volume, $$\Delta {U_{298\,K}} = - 240.60\,kJ\,mo{l^{ - 1}}$$ and $${C_V}$$ values for $${H_2}O$$ vapour and $${N_2}$$ in the temperature range $$298\,K$$ and $$3200\,K$$ are $$39.06\,J{K^{ - 1}}\,mo{l^{ - 1}}$$ and $$26.40\,J{K^{ - 1}}\,mo{l^{ - 1}}$$ respectively. The explosion temperature under adiabatic conditions is ( Given : $${n_{{N_2}}} = 2$$ )
A.
$$2900\,K$$
B.
$${2900^ \circ }C$$
C.
$$2917\,K$$
D.
$${3000^ \circ }C$$
Answer :
$$2917\,K$$
Solution :
If the process is carried out adiabatically and isochorically,
$$\eqalign{
& \Delta U = \Delta {U_{{\text{heating}}}} + \Delta {U_{298\,K}} = 0\,\,{\text{or}}\,\,\Delta {U_{{\text{heating}}}} = - \Delta {U_{298\,K}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \int\limits_{298\,K}^{{T_f}} {n\sum {{C_\nu }dT} = + 240.60\,kJ\,mo{l^{ - 1}}} \cr
& \sum {n{C_\nu } = n \cdot {C_{\nu \left( {{H_2}{O_{\left( v \right)}}} \right)}} + n{C_{\nu \left( {{N_{2\left( g \right)}}} \right)}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {39.06 + 2 \times 26.40} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 91.86\,J{K^{ - 1}}\,mo{l^{ - 1}} \cr} $$
by using the value of $$\sum {n{C_\nu }} $$ in the above equation
$$\eqalign{
& \left( {91.86} \right)\left( {{T_f} - 298} \right) = 240600\,J\,mo{l^{ - 1}} \cr
& {T_f} - 298 = \frac{{240600}}{{91.86}} = 2619\,K \cr
& {T_f} = 2619 + 298 = 2917\,K \cr} $$