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Igoryamba
3 years ago
11

Consider the reaction data. A ⟶ products T ( K ) k ( s − 1 ) 225 0.385 525 0.635 What two points should be plotted to graphicall

y determine the activation energy of this reaction? To avoid rounding errors, use at least three significant figures in all values. x 1 = y 1 = x 2 = y 2 = Determine the rise, run, and slope of the line formed by these points. rise: run: slope: What is the activation energy of this reaction? E a = J / mol

Physics
1 answer:
lutik1710 [3]3 years ago
4 0

Answer:

Plot ln K vs 1/T

(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol

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 = - \dfrac{E_{a}}{R}\dfrac{1}{T} + \ln A\\\\y = mx + b

Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA

Data:

\begin{array}{cccc}\textbf{k/s}\mathbf{^{-1}} &\mathbf{\ln k} & \textbf{T/K} & \mathbf{1/T(K^{-1})}\\0.285 & -0.9545 & 225 &0.004444\\0.635 & -0.4541 & 525 & 0.001905\\\end{array}

Calculations:

(a) Rise

Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004

(b) Run

Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹

(c) Slope

Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹

(d) 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

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