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Nernst Equation

For first ord...

Question

For first order parallel reaction $k_{1}$ and $k_{2}$ are $4$ and $2$ ${m i n}^{- 1}$ respectively at $300 K$ . If the activation energies for the formation of $B$ and $C$ are respectively $30, 000$ and $38, 314 j o u l e / m o l$ respectively. The temperature at which $B$ and $C$ will be obtained in equimolar ratio is:

A

757.48 K

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B

378.74 K

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C

600 K

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D

none of these

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Solution

The correct option is B $378.74 K$
Two equations will be useful $A = A_{0} E^{- k t}$ and $\frac{k_{2}}{k_{1}} = e^{- \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$ .
For the product B, $k = 2 e^{- \frac{30000}{8.314} (\frac{T_{2} - 300}{300 T_{2}})}$ .
And for the product C, $k = e^{- \frac{38314}{8.314} (\frac{T_{2} - 300}{300 T_{2}})}$ .
When equimolar mixture of B and C is formed,
$2 e^{- \frac{30000}{8.314} (\frac{T_{2} - 300}{300 T_{2}})} = e^{- \frac{38314}{8.314} (\frac{T_{2} - 300}{300 T_{2}})}$
Hence, $T_{2} = 378.74 K$ .

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