Question

Consider the reaction data.

Answer 1

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:

$k = Ae^{-E_{a}/RT}\\\text{Take the ln of each side}\\\ln k = \ln A - \dfrac{E_{a}}{RT}\\\\\text{We can rearrange this to give}\\\ln k = -\left ( \dfrac{E_{a}}{R} \right )\dfrac{1}{T} + \ln A$

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