The rate constants k1, and k2, for two different reactions are 1016⋅e−2000/T and 1015⋅e−1000/Trespectively. The temperature atwhich k1=k2 is

Given,k1=1016⋅e−2000Tand k2=1015⋅e−1000TWhen k1 and k2are equal at any temperature T, we have 1016⋅e−2000T=1015⋅e−1000Tor 10×1015⋅e−2000T=1015⋅e−1000Tor 10.e−2000T=e−1000Tor ln⁡10−2000T=−1000Tor ln⁡10=2000T−1000Tor 2.303log⁡10=1000Tor 2.303×1×T=1000 [∴log⁡10=1]or T=10002.303K

The rate constants k1, and k2, for two different reactions are 1016⋅e−2000/T and 1015⋅e−1000/Trespectively. The temperature atwhich k1=k2 is

Given,k1=1016⋅e−2000Tand k2=1015⋅e−1000TWhen k1 and k2are equal at any temperature T, we have 1016⋅e−2000T=1015⋅e−1000Tor 10×1015⋅e−2000T=1015⋅e−1000Tor 10.e−2000T=e−1000Tor ln⁡10−2000T=−1000Tor ln⁡10=2000T−1000Tor 2.303log⁡10=1000Tor 2.303×1×T=1000 [∴log⁡10=1]or T=10002.303K

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The rate constants k₁, and k₂, for two different reactions are $10^{16} \cdot e^{- 2000 / T} and 10^{15} \cdot e^{- 1000 / T}$ respectively. The temperature $a t which k_{1} = k_{2} i s$

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1000K

$\frac{2000}{2 .303} K$

2000K

$\frac{1000}{2 .303} K$

answer is D.

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Detailed Solution

Given, $k_{1} = 10^{16} \cdot e^{- \frac{2000}{T}}$
and $k_{2} = 10^{15} \cdot e^{- \frac{1000}{T}}$

When $k_{1} and k_{2}$ are equal at any temperature T, we have
$10^{16} \cdot e^{- \frac{2000}{T}} = 10^{15} \cdot e^{- \frac{1000}{T}}$
or $10 \times 10^{15} \cdot e^{- \frac{2000}{T}} = 10^{15} \cdot e^{- \frac{1000}{T}}$
or $10 . e^{- \frac{2000}{T}} = e^{- \frac{1000}{T}}$
or $\ln 10 - \frac{2000}{T} = - \frac{1000}{T}$
or $\ln 10 = \frac{2000}{T} - \frac{1000}{T}$
or $2 .303 \log 10 = \frac{1000}{T}$
or $2 .303 \times 1 \times T = 1000 [∴ \log 10 = 1]$
or $T = \frac{1000}{2 .303} K$