Question
Rate constant $$k = 1.2 \times {10^3}mo{l^{ - 1}}\,L\,{s^{ - 1}}$$ and $${E_a} = 2.0 \times {10^2}kJ\,mo{l^{ - 1}}.$$ When $$T \to \infty :$$
A.
$$A = 2.0 \times {10^2}kJ\,mo{l^{ - 1}}$$
B.
$$A = 1.2 \times {10^3}\,mo{l^{ - 1}}\,L\,{s^{ - 1}}$$
C.
$$A = 1.2 \times {10^3}mol\,{L^{ - 1}}{s^{ - 1}}$$
D.
$$A = 2.4 \times {10^3}kJ\,mo{l^{ - 1}}{s^{ - 1}}$$
Answer :
$$A = 1.2 \times {10^3}\,mo{l^{ - 1}}\,L\,{s^{ - 1}}$$
Solution :
$$\eqalign{
& k = A{e^{ - \,\frac{{Ea}}{{RT}}}} \cr
& {\text{If}}\,\,T \to \infty ,\,k = A \cr} $$