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The rate constant k , for the reaction `N_(2)O_(5) to 2NO_(2) (g) + (1)/(2) O_(2)(g)` correspond to concentration of `N_(2)O_(5)` initially and at time t .A. `[N_(2)O_(5)]_(t) [N_(2)O_(5)]_(0) + kt`B. `[N_(2)O_(5)]_(0) = [N_(2)O_(5)]_(t) e^(kt)`C. `log_(10) [N_(2)O_(5)]_(t) = log_(10)[N_(2)O_(5)]_(0) - kt`D. ln `([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))= kt` |
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Answer» Correct Answer - d Rate constant = `2.3 xx 10^(-2) sec^(-1)` It means it is a first order reaction (because unit of rate constant is `sec^(-1)`) For first order reaction k = `(1)/(t)` ln `(a)/(a-x)` Kt = ln `(a)/(a-x)` = ln `([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))`. |
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