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Two reactions...

Question

Two reactions $A$ and $B$ of the same order have equal pre-exponential factors but activation energy for $A$ is greater than that of $B$ by $24.9 k J / m o l$ . Calculate the ratio( $\frac{k_{B}}{k_{A}}$ ) at $27^{\circ} C .$
Here, $k_{A}$ and $k_{B}$ are rate constants for reactions $A$
and $B$ , respectively.
$(R = 8.31 J K^{- 1} m o l^{- 1})$

2.199 \times 10^{4}

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2.199 \times 10^{- 4}

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4.199 \times 10^{- 4}

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4.199 \times 10^{4}

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Solution

The correct option is A $2.199 \times 10^{4}$
We know that,
$l o g_{10} k = l o g_{10} A - \frac{E}{2.303 R T}$
So, $l o g_{10} k_{B} = l o g_{10} A - \frac{E_{B}}{2.303 R T}$
and $l o g_{10} k_{A} = l o g_{10} A - \frac{E_{A}}{2.303 R T}$
$l o g_{10} (\frac{k_{B}}{k_{A}}) = \frac{(E_{A} - E_{B})}{2.303 R T}$
$l o g_{10} (\frac{k_{B}}{k_{A}}) = \frac{24.9 \times 1000}{2.303 \times 8.314 \times 300}$
$\frac{k_{B}}{k_{A}} = 2.199 \times 10^{4}$
Hence, (a) is correct.

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Two reactions A and B of the same order have equal pre-exponential factors but activation energy for A is greater than that of B by 24.9 kJ/mol. Calculate the ratio(kBkA) at 27∘C. Here, kA and kB are rate constants for reactions A and B, respectively. (R=8.31 JK−1mol−1)

The correct option is A 2.199×104We know that, log10k=log10A−E2.303RT So, log10kB=log10A−EB2.303RT and log10kA=log10A−EA2.303RT log10(kBkA)=(EA−EB)2.303RT log10(kBkA)=24.9×10002.303×8.314×300 kBkA=2.199×104 Hence, (a) is correct.