For a first-order chemical reaction, the rate constant k follows the Arrhenius equation. Which of the following correctly describes the plot of the natural logarithm of its half-life ( t_ 1/2 ) versus 1 T ?

Step-by-step solution

For a first-order reaction, the half-life is given by: t_ 1/2 = 0.693 k According to the Arrhenius equation, k = A e^ -E_a/RT . Substituting this into the half-life expression gives: t_ 1/2 = 0.693 A e^ -E_a/RT = ( 0.693 A ) e^ E_a/RT Taking the natural logarithm on both sides, we get: t_ 1/2 = ( 0.693 A ) + E_a RT This is in the form of a straight line equation, y = mx + c , where y = t_ 1/2 and x = 1 T . The slope m = E_a R , which is a positive value. The intercept c = ( 0.693 A ) , which means the line does not