1.

For a chemical reaction at 27^(@)C , the activation energy is 600 R . The ratio of the rate constants at 327^(@)C to that of at 27^(@)C will be

Answer»

2
40
E
`e^(2)`

Solution :LN `(K_(2))/(K_(1)) = (E_(a))/(R) ((1)/(T_(1)) - (1)/(T_(2)))` or , ln `(K_(2))/(K_(1)) = (600R)/(R) ((1)/(300) - (1)/(600))`
or , ln `(K_(2))/(K_(1)) = (600R)/(R) ((2-1)/(600)) = 1 `
ln `(K_(2))/(K_(1))` = ln e `"" (K_(2))/(K_(1)) = e`


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