NEETChemistryChemical Kinetics
Match List-I with List-II. List-I List-II (A) Slope of k vs 1 T (I) ₁₀ A (B) Intercept of k vs 1 T (II) - E_a 2.303 R (C) Slope of ₁₀ k vs 1 T (III) A (D) Intercept of ₁₀ k vs 1 T (IV) - E_a R Choose the correct answer from the options given below:
Options
- A(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- B(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
- C(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- D(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Correct answer
A. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Step-by-step solution
The Arrhenius equation is k = A e^ - E_a RT . Taking the natural logarithm on both sides yields: k = A - E_a RT Comparing this with the equation of a straight line y = mx + c , a plot of k vs 1 T has a slope of - E_a R and an intercept of A . Converting the equation to base 10 logarithm gives: 2.303 ₁₀ k = 2.303 ₁₀ A - E_a RT ₁₀ k = ₁₀ A - E_a 2.303 RT Comparing this with y = mx + c , a plot of ₁₀ k vs 1 T has a slope of - E_a 2.303 R and an intercept of ₁₀ A . Therefore, the correct matches are: (A) Slope of k vs