Answer:
(a) x₁= 0.004 444; (b) y₁ = -0.9545; (c) x₂ = 0.001 905; (d) y₂ = -0.4541;
(e) rise = 0.5004; (f) run = -0.002 539; (g) slope = -197.1; (h) Eₐ = -1.64 kJ·mol⁻¹
Explanation:
This is an example of the Arrhenius equation:
Thus, if you plot ln k vs 1/T, you should get a straight line with slope = -Eₐ/R and a y-intercept = lnA
(a) x₁
x₁= 1/T₁ = 1/225 = 0.004 444
(b) y₁
y₁ = ln(k₁) = ln0.385 = -0.9545
(c) x₂
x₂= 1/T₂ = 1/525 = 0.001 905
(d) y₂
y₂ = ln(k₂) = ln0.635 = -0.4541
(e) Rise
Δy = y₂ - y₁ = -0.4541 - (-0.9545) = -0.4541 + 0.9545 = 0.5004
(f) Run
Δx = x₂ - x₁ = 0.001 905 - 0.004 444 = -0.002 539
(g) Slope
Δy/Δx = 0.5004/(-0.002 539) K⁻¹ = -197.1
(h) Activation energy
Slope = -Eₐ/R
Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol
