Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be A. `2.303 xx 2 cal`B. `2//2.303 cal`C. `2 cal`D. None

Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T`

1.	Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be A. `2.303 xx 2 cal`B. `2//2.303 cal`C. `2 cal`D. None
Answer» Correct Answer - C `logk=logA-(E_(a))/(2.303 RT)(y=c+mx)` Slope `=(-E_(a))/(2.303R)=(1)/(2.303)(given)(tan theta=(1)/(2.303))` `-E_(a)= 2.303Rxxslope` `=2.303xx(R )/(2.303)=R=2 cal`

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Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be A. `2.303 xx 2 cal`B. `2//2.303 cal`C. `2 cal`D. None

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