Question

Rate of disappearance of the reactant A in the reversable and one step reaction $A\rightleftharpoons B$ at two temperatures is given by
$\frac{-d\left[A\right]}{dt}=2.0\times {10}^{-3}{s}^{-1}\left[A\right]-5.0\times {10}^{-4}{s}^{-1}\left[B\right]\left(at{27}^{\circ }C\right)\frac{-d\left[A\right]}{dt}=8.0\times {10}^{-2}{s}^{-1}\left[A\right]-4.0\times {10}^{-3}{s}^{-1}\left[B\right]\left(at{127}^{\circ }C\right)$
The enthapy of the reaction in the given temperature range is

• A. $-\frac{2.303\times 8.314\times 300\times 400}{100}log5J$
• B. $\frac{2.303\times 8.314\times 300\times 400}{100}log50J$
• C. $\frac{2.303\times 8.314\times 300\times 400}{100}log5J$
• D. $\frac{100}{2.303\times 8.314\times 300\times 400}log5J$

Answer 1

The correct option is C

$\frac{2.303\times 8.314\times 300\times 400}{100}log5J$

$K_{1}at{27}^{\circ }C=\frac{2.0\times {10}^{-3}}{5\times {10}^{-4}}=4.0K_{2}at{127}^{\circ }C=\frac{8\times {10}^{-2}}{4.0\times {10}^{-3}}=20log\frac{K_2}{K_1}=\frac{\Delta H}{2.303R}\frac{[T_2-T_1]}{T_1{T}_2}{T}_1=300K,T_2=400K$