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Standard XII

Chemistry

Arrhenius Equation

Two reactions...

Question

Two reactions of the same order have equal exponential factors but their activation energies differ by $24.9 k J / m o l$ . Calculate the ratio between the rate constants of these reactions at $27^{o} C$ . $(R = 8.31 J K^{- 1} {m o l}^{- 1})$

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Solution

We know that,
${log}_{10} k = {log}_{10} A - \frac{E}{2.303 R T}$
So, ${log}_{10} k_{2} = {log}_{10} A - \frac{E_{2}}{2.303 R T}$
and ${log}_{10} k_{1} = {log}_{10} A - \frac{E_{1}}{2.303 R T}$
${log}_{10} (\frac{k_{2}}{k_{1}}) = \frac{(E_{1} - E_{2})}{2.303 R T}$
${log}_{10} (\frac{k_{2}}{k_{1}}) = \frac{24.9 \times 1000}{2.303 \times 8.314 \times 300}$
$\frac{k_{2}}{k_{1}} = 2.162 \times 10^{4}$

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Two reactions of the same order have equal exponential factors but their activation energies differ by $24.9 k J / m o l$ . Calculate the ratio of the rate constants of this reaction at $27^{\circ} C$ . $(R = 8.314 J K^{- 1} m o l^{- 1})$

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Two reactions of the same order have equal exponential factors but their activation energies differ by 24.9 kJ/mol. Calculate the ratio between the rate constants of these reactions at 27oC. (R=8.31 J K−1mol−1)

We know that,log10k=log10A−E2.303RTSo, log10k2=log10A−E22.303RTand log10k1=log10A−E12.303RTlog10(k2k1)=(E1−E2)2.303RTlog10(k2k1)=24.9×10002.303×8.314×300k2k1=2.162×104

Two reactions of the same order have equal exponential factors but their activation energies differ by $24.9 k J / m o l$ . Calculate the ratio between the rate constants of these reactions at $27^{o} C$ . $(R = 8.31 J K^{- 1} {m o l}^{- 1})$