A) \[\frac{2000}{2.303}K\]
B) 2000 K
C) \[\frac{1000}{2.303}K\]
D) 1000 K
Correct Answer: C
Solution :
\[{{K}_{1}}={{10}^{16}}{{e}^{-2000/T}};{{K}_{2}}=1015\,{{e}^{-1000/T}}\] if \[{{K}_{1}}={{K}_{2}}\]then \[{{10}^{16}}{{e}^{-2000/T}}={{10}^{15}}{{e}^{-1000/T}}\] or \[log10-\frac{2000}{T}=-\frac{1000}{T}\]or \[T=\frac{1000}{2.303}K\]You need to login to perform this action.
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