Question
Rate of disappearance of the reactant ‘A’ in the reversible reaction A$\rightleftharpoons$B at two temperatures is given as $- \frac{d\lbrack A\rbrack}{dt}$= (2.0 × 10–3 s–1) [A] – (5.0 × 10–4 s–1) [B] (at 27°C)
$- \frac{d\lbrack A\rbrack}{dt}$= (8.0 × 10–2 s–1) [A] – (4.0 × 10–3 s–1) [B] (at 127°C)
The enthalpy of reaction in the given temperature range is
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Solution
Keq at 27°C, $K_{1} = \frac{2 \times 10^{- 3}}{2 \times 10^{- 4}} = 4$
and Keq at 127°C, $K_{2} = \frac{8 \times 10^{- 2}}{4 \times 10^{- 3}} = 20$
Now, $\ln\frac{K_{2}}{K_{1}} = \frac{\Delta H}{R}\left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$
Or, $2.303\log\frac{20}{4} = \frac{\Delta H}{R}\left( \frac{1}{300} - \frac{1}{400} \right)$
∴ ΔH = 2.303 × 8.314 × 1200 log(5) J/mol
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