The plot of $\log _ { 10 } \mathrm { Kvs } \frac { 1 } { \mathrm {~T} }$ gives a straight line. The intercept and slope respectively are
(A) $\mathrm { c } = \underline { \log \mathrm { A } } , \mathrm { m } = - \frac { \mathrm { E } _ { \mathrm { a } } } { 2.303 \mathrm { R } }$
(B) $c = - \frac { \mathrm { E } _ { q } } { 2,303 \mathrm { R } ^ { \prime } } , \mathrm { m } = \log \mathrm { A }$
(C) $c = -$ Iog $\mathrm { A } , \mathrm { m } = - \frac { \mathrm { E } _ { \mathrm { a } } } { 2.303 \mathrm { R } }$
(D) $c = \underline { \log A } , m = \frac { E _ { n } } { 2.303 R }$