MCQ
Temperature dependent equation can be written as
- ✓$\ln \,k = \ln A - {e^{{E_a}/RT}}$
- B$\ln k = \ln A + {e^{{E_a}/RT}}$
- C$\ln k = \ln A - {e^{RT/{E_a}}}$
- DAll of these
$K = A{e^{ - {E_a}/RT}}$
$\ln K = \ln A - {e^{{E_a}/RT}}$
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Element $-$ $IP$
The time taken for $A$ to become $1 / 4^{\text {th }}$ of its inital concentration is twice the time taken to become $1 / 2$ of the same. Also, when the change of concentration of $B$ is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is . . . . .