Problem 1

The gas-phase isomerisation between cis-but-2-ene and trans-but-2-ene is reversible and unimolecular, with equilibrium constant

K

and forward/reverse rate constants

k_1

and

k_{-1}

\ce{cis-but-2-ene <=>[

k_1

][

k_{-1}

] trans-but-2-ene}

. Initially the system contains only cis-but-2-ene at concentration

C_s^0

. (a) Write the differential rate equation for

C_s^t

. (b) Derive the expression for the concentration ratio

C_s^t/C_s^0

as a function of

t

K

and

k_{-1}

. (c) When

K = 1

, identify which of the following linear plots is valid: (H-1)

\ln(C_s^t - 0.5C_s^0)

t

; (H-2)

\ln(C_s^t)

t

; (H-3)

1/\ln(C_s^t - 0.5C_s^0)

t

; (H-4)

1/(C_s^t - 0.5C_s^0)

t

. (d) At

T = 690\ \text{K}

K = 1.14

and

k_1 = 1.6 \times 10^{-6}\ \text{s}^{-1}

. Calculate the time (in hours) required for cis-but-2-ene to convert to 30 % of its maximum equilibrium conversion. (e) Calculate the percentage of trans-but-2-ene in the mixture after 10 hours.

Step 5 of 5: Composition after 10 hours

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t = 36000\ \text{s} \Rightarrow \frac{C_s^t}{C_s^0} = \frac{1}{2.14} + \frac{1.14}{2.14}\exp[-1.404 \times 10^{-6} \times 36000 \times 2.14] = 0.9454 \Rightarrow \%\,\text{trans} = 5.46\,\%

Analysis

Substitute $t = 10\ \text{h} = 36000$ s into the integrated equation: $C_s^t/C_s^0 = 0.9454$ , so $C_t^t/C_s^0 = 1 - 0.9454 = 0.0546$ , meaning 5.46 % trans isomer — exactly matching the official key.