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:
[tex]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[/tex]
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
(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
It is given by:
[tex]k=Ae^{\frac{-E_a}{RT}}[/tex]
On taking the log both sides, equation will be:
[tex]ln k = -(\frac{-E_a}{RT})\frac{1}{T} +ln A[/tex]
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
Find more information about Activation energy here:
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