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ChemHardAdv 2023 • Paper 1

Calculation of Backward Rate Constant Logarithm at Given Temperature

The plot of log⁡kf\log k_f versus 1/T1/T for a reversible reaction A (g)⇌P (g)\text{A (g)} \rightleftharpoons \text{P (g)} is shown.

Pre-exponential factors for the forward and backward reactions are 1015 s−110^{15}\text{ s}^{-1} and 1011 s−110^{11}\text{ s}^{-1}, respectively. If the value of log⁡K\log K for the reaction at 500 K500\text{ K} is 66, the value of ∣log⁡kb∣|\log k_b| at 250 K250\text{ K} is _____.

[K=equilibrium constant of the reactionK = \text{equilibrium constant of the reaction} kf=rate constant of forward reactionk_f = \text{rate constant of forward reaction} kb=rate constant of backward reactionk_b = \text{rate constant of backward reaction}]

Question Diagram 1
Official Numerical Answer5

Topics & Concepts

Chemical KineticsArrhenius EquationRate of Reaction

Step-by-Step Solution

To find the value of ∣log⁡kb∣|\log k_b| at T=250 KT = 250\text{ K}, we use the Arrhenius equation for both the forward and backward reactions.

Step 1: Calculate the Activation Energy of the Forward Reaction (Ea,fE_{a,f})

The Arrhenius equation in logarithmic form (base 1010) for the forward reaction is: log⁡kf=log⁡Af−Ea,f2.303RT\log k_f = \log A_f - \frac{E_{a,f}}{2.303 R T}

Given:

  • Af=1015 s−1  ⟹  log⁡Af=15A_f = 10^{15}\text{ s}^{-1} \implies \log A_f = 15
  • From the graph, at 1T=0.002 K−1\frac{1}{T} = 0.002\text{ K}^{-1} (which corresponds to T=500 KT = 500\text{ K}), the value of log⁡kf=9\log k_f = 9.

Substituting these values into the forward rate constant equation: 9=15−Ea,f2.303R(0.002)9 = 15 - \frac{E_{a,f}}{2.303 R} (0.002)

Ea,f2.303R×0.002=15−9=6\frac{E_{a,f}}{2.303 R} \times 0.002 = 15 - 9 = 6

Ea,f2.303R=60.002=3000 K\frac{E_{a,f}}{2.303 R} = \frac{6}{0.002} = 3000\text{ K}

Step 2: Calculate the Backward Rate Constant (log⁡kb\log k_b) at T=500 KT = 500\text{ K}

For a reversible reaction, the equilibrium constant KK is given by: K=kfkb  ⟹  log⁡K=log⁡kf−log⁡kbK = \frac{k_f}{k_b} \implies \log K = \log k_f - \log k_b

At T=500 KT = 500\text{ K}:

  • log⁡K=6\log K = 6
  • log⁡kf=9\log k_f = 9

Substituting these values: 6=9−log⁡kb(500 K)6 = 9 - \log k_b (500\text{ K}) log⁡kb(500 K)=9−6=3\log k_b (500\text{ K}) = 9 - 6 = 3

Step 3: Calculate the Activation Energy of the Backward Reaction (Ea,bE_{a,b})

The logarithmic form of the Arrhenius equation for the backward reaction is: log⁡kb=log⁡Ab−Ea,b2.303RT\log k_b = \log A_b - \frac{E_{a,b}}{2.303 R T}

Given:

  • Ab=1011 s−1  ⟹  log⁡Ab=11A_b = 10^{11}\text{ s}^{-1} \implies \log A_b = 11
  • At T=500 KT = 500\text{ K} (1T=0.002 K−1\frac{1}{T} = 0.002\text{ K}^{-1}), log⁡kb=3\log k_b = 3.

Substituting these values: 3=11−Ea,b2.303R(0.002)3 = 11 - \frac{E_{a,b}}{2.303 R} (0.002)

Ea,b2.303R×0.002=11−3=8\frac{E_{a,b}}{2.303 R} \times 0.002 = 11 - 3 = 8

Ea,b2.303R=80.002=4000 K\frac{E_{a,b}}{2.303 R} = \frac{8}{0.002} = 4000\text{ K}

Step 4: Calculate ∣log⁡kb∣|\log k_b| at T=250 KT = 250\text{ K}

At T=250 KT = 250\text{ K}, we have: 1T=1250=0.004 K−1\frac{1}{T} = \frac{1}{250} = 0.004\text{ K}^{-1}

Using the Arrhenius equation for the backward reaction at T=250 KT = 250\text{ K}: log⁡kb(250 K)=log⁡Ab−Ea,b2.303R(1T)\log k_b (250\text{ K}) = \log A_b - \frac{E_{a,b}}{2.303 R} \left(\frac{1}{T}\right)

log⁡kb(250 K)=11−(4000)×(0.004)\log k_b (250\text{ K}) = 11 - (4000) \times (0.004)

log⁡kb(250 K)=11−16=−5\log k_b (250\text{ K}) = 11 - 16 = -5

Taking the absolute value: ∣log⁡kb∣=∣−5∣=5|\log k_b| = |-5| = 5

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