A plot of log k against  \(\frac { 1 }{ T } \) values gives a straight line with slope value equal to - E_a/ 2.303R and intercept value equals to log A.

(ii) The plot gives a negative slope straight line also.
(iii) From the slope of straight line, E_a can be calculated.

Average rate = \(-\frac { 1 }{ 2 } \frac { \triangle \left[ A \right] }{ \triangle L } \)
\(=-\frac { 1 }{ 2 } \frac { { \left[ A \right] }_{ 2 }-{ \left[ A \right] }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } \)
\(=-\frac { 1 }{ 2 } \left( \frac { 0.4-0.5 }{ 10 } \right) \)
\(=-\frac { 1 }{ 2 } \left( \frac { -0.1 }{ 10 } \right) \)
= 0.005 mol L^-1 min^-1
= 5 x 10^-3 M min^-1

\({ k }=2.303\log { \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } } \)
\(t=\frac { 2.303 }{ { 10 }^{ -3 } } \log { \frac { 10 }{ 2.5 } } \)
t = 2303 x 0.301 x 2 = 1386 s.

(i) Catalyst lowers the activation energy and changes the path of the reaction and thus increases the rate of the reaction. A catalyst does not alter Gibbs energy of a reaction.
(ii) Higher the temperature faster the reaction. This is because increasing the temperature increases the Kinetic energy, and hence more molecules would have energy greater than the activation energy resulting in increased rate and completion of the reaction. Hence oxalic acid should be heated in redox titration against KMnO₄.

For n^th reaction
\(K\alpha \frac { 1 }{ { C }^{ n-1 } } \)