A) 1000 K
B) \[\frac{2000}{2.303}K\]
C) 2000 K
D) \[\frac{1000}{2.303}K\]
Correct Answer: D
Solution :
Key Idea: The Arrhenius equation is represented as\[k=A{{e}^{-{{E}_{a}}/RT}}\] In the given equations, first take log and. Then compare them. \[{{k}_{1}}={{10}^{16}}{{e}^{-2000/T}}\] \[{{k}_{2}}={{10}^{15}}{{e}^{-1000/T}}\]On taking log, we get \[\log {{k}_{1}}=\log {{10}^{16}}-\frac{2000}{2.303T}\] ?(i) \[\log {{k}_{2}}=\log {{10}^{15}}-\frac{1000}{2.303T}\] ?(ii) \[\because \]\[{{k}_{1}}={{k}_{2}}\]Hence, from Eqs (i) and (ii) \[T=\frac{1000}{2.303}K\]
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