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  • For gaseous reaction, following data is given: <img alt="\mathrm{ A \rightarrow B, k_1=1015 e^{-2000 / T} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20A%20%5Crighta

For gaseous reaction, following data is given:
\mathrm{ A \rightarrow B, k_1=1015 e^{-2000 / T} }
\mathrm{ C \rightarrow D, k_2=1014 e^{-1000 / T}}
The temperature at which \mathrm{\mathrm{k}_1=\mathrm{k}_2} is:

Option: 1

1000 K


Option: 2

2000 K


Option: 3

868.8 K


Option: 4

434.2 K


Answers (1)

best_answer

According to Arrhenius equation,
\mathrm{k=A e^{-E_a / R T}}
where,
k : Rate constant
A : Arrhenius factor or frequency factor or pre-exponential factor
\mathrm{E_a:} Activation energy (in $\mathrm{J} / \mathrm{mol}$ )
R : Gas constant
T : Temperature of reaction (in Kelvin(K))
When \mathrm{k_1=k_2}
\mathrm{ 10^{15} e^{-2000 / T}=10^{14} e^{-1000 / T} }

\mathrm{ 10=e^{-1000 / T} e^{2000 / T} \\ }

\mathrm{ 10=e^{\left(\frac{-1000}{T}+\frac{2000}{T}\right)} \\ }

\mathrm{ 10=e^{1000 / T} }
Taking ln on both sides
\mathrm{ \Rightarrow \ln 10=\frac{1000}{T} \\ }

\mathrm{ \Rightarrow 2.303 \log 10=\frac{1000}{T} \\ }

\mathrm{ \Rightarrow T=\frac{1000}{2.303} }
Hence, \mathrm{ T=434.2 \mathrm{~K} }

Posted by

Ritika Harsh

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