For a chemical reaction, the temperature dependence of the rate constant k is studied by plotting k versus 1 T . If a positive catalyst is added to the reaction (assuming the pre-exponential factor remains constant), how will the resulting straight line change compared to the uncatalysed reaction?

Step-by-step solution

According to the Arrhenius equation, k = A - E_a RT . When k is plotted against 1 T , a straight line is obtained with a slope equal to - E_a R and a y-intercept equal to A . The addition of a positive catalyst provides an alternate pathway with a lower activation energy ( E_a ). Since the new E_a is smaller, the term - E_a R becomes less negative (its magnitude decreases). Assuming the pre-exponential factor A remains constant, the y-intercept is unchanged, but the slope becomes less negative, resulting in a flatt