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 3 of 5: Identification of linear plot

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K = 1 \Rightarrow \ln(C_s^t - 0.5C_s^0) = \ln(0.5C_s^0) - 2k_{-1}t \Rightarrow \text{Graph H-1 is linear}

Analysis

Setting $K = 1$ reduces the integrated law to $C_s^t - 0.5C_s^0 = 0.5C_s^0 \exp(-2k_{-1}t)$ , taking natural logarithms gives $\ln(C_s^t - 0.5C_s^0) = \ln(0.5C_s^0) - 2k_{-1}t$ , which matches graph H-1.