Question

Rate of disappearance of the reactant A at two different temperature is given by $A\rightleftharpoons B$.
$\frac{-d\left[A\right]}{dt}=(2\times {10}^{-2}{S}^{-1})\left[A\right]-4\times {10}^{-3}{S}^{-1}\left[B\right];300K$
$\frac{-d\left[A\right]}{dt}=(4\times {10}^{-2}{S}^{-1})\left[A\right]-16\times {10}^{-4}{S}^{-1}\left[B\right];400K$
Calculate heat of reaction in the given temperature range, when equilibrium is set up.

• A. $8.03$ kJ
• B. $16.06$ kJ
• C. $32.12$ kJ
• D. None of these

Answer 1

The correct option is B $16.06$ kJ

At 300 K; $K_1=\frac{2\times {10}^{-2}}{4\times {10}^{-3}}=5$
At 400 K; $K_2=\frac{4\times 1{-}^{-2}}{16\times {10}^{-4}}=25$
$log\frac{K_2}{K_1}=\frac{\Delta H}{2.303\times 8.314}[\frac{1}{300}-\frac{1}{400}]$
So, change in heat of reaction is 16.06 kJ.