Take $R = 8.314 J mo l^{- 1} K^{- 1}$ . An exothermic reaction has $Δ H^{\circ} = - 92.0 kJ mo l^{- 1}$ (constant). Its equilibrium constant at $450 K$ is $K_{1}$ . At what temperature (in K, nearest integer) is the equilibrium constant equal to $0.15 K_{1}$ ?

Numerical answer type, answer in K, tolerance 1.

Answer: 488 K

From

ln (K_{2} / K_{1}) = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})

we get

\frac{1}{T _{2}} = \frac{1}{T _{1}} - \frac{R ln ( K _{2} / K _{1} )}{Δ H ^{\circ}} = \frac{1}{450} - \frac{( 8.314 ) ( - 1.8971 )}{- 92000} = 2.0508 \times 1 0^{- 3} K^{- 1}

. Hence

T_{2} = 487.62 K

, i.e. 488 K. It is higher than

450 K

because

K

of an exothermic reaction falls on heating.