Question

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

• A. $16.06kJ$
• B. $23.04kJ$
• C. $26.78kJ$
• D. $29.34kJ$

Answer 1

Answer: (A)

$16.06kJ$

At $300K$, the equilibrium constant is $K=\frac{2\times {10}^{-2}}{4\times {10}^{-3}}=5$

At $400K$, the equilibrium constant is $K^{\prime }=\frac{4\times {10}^{-2}}{16\times {10}^{-4}}=25$

$ln\frac{K^{\prime }}{K}=\frac{\Delta H}{R}\frac{T^{\prime }-T}{T{T}^{\prime }}$

$ln\frac{25}{5}=\frac{\Delta H}{8.314}\frac{400-300}{300\times 400}$

$\Delta H=16060J$

$=16.06kJ$