Chemistry Labs

Problem 1

The gas-phase isomerisation between cis-but-2-ene and trans-but-2-ene is reversible and unimolecular, with equilibrium constant KK and forward/reverse rate constants k1k_1 and k−1k_{-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 Cs0C_s^0. (a) Write the differential rate equation for CstC_s^t. (b) Derive the expression for the concentration ratio Cst/Cs0C_s^t/C_s^0 as a function of tt, KK and k−1k_{-1}. (c) When K=1K = 1, identify which of the following linear plots is valid: (H-1) ln⁡(Cst−0.5Cs0)\ln(C_s^t - 0.5C_s^0) vs tt; (H-2) ln⁡(Cst)\ln(C_s^t) vs tt; (H-3) 1/ln⁡(Cst−0.5Cs0)1/\ln(C_s^t - 0.5C_s^0) vs tt; (H-4) 1/(Cst−0.5Cs0)1/(C_s^t - 0.5C_s^0) vs tt. (d) At T=690 KT = 690\ \text{K}, K=1.14K = 1.14 and k1=1.6×10−6 s−1k_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 4 of 5: Time for 30 % of maximum conversion
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k−1=1.6×10−61.14=1.404×10−6 s−1;exp⁡[−k−1t(K+1)]=0.70⇒t=32.98 hk_{-1} = \frac{1.6 \times 10^{-6}}{1.14} = 1.404 \times 10^{-6}\ \text{s}^{-1};\quad \exp[-k_{-1}t(K+1)] = 0.70 \Rightarrow t = 32.98\ \text{h}
Analysis

Maximum conversion is Cs0−Cs,eq=KCs0K+1C_s^0 - C_{s,\text{eq}} = \frac{K C_s^0}{K+1}. At 30 % conversion, Cst=Cs0−0.3KCs0K+1C_s^t = C_s^0 - 0.3\frac{K C_s^0}{K+1}, so KK+1exp⁡[−k−1t(K+1)]=0.7KK+1\frac{K}{K+1}\exp[-k_{-1}t(K+1)] = 0.7\frac{K}{K+1}, which simplifies to exp⁡[−k−1t(K+1)]=0.7\exp[-k_{-1}t(K+1)] = 0.7. Then t=−ln⁡(0.7)/[(1.14+1)(1.404×10−6)]=118712 s=32.98 ht = -\ln(0.7)/[(1.14+1)(1.404 \times 10^{-6})] = 118712\ \text{s} = 32.98\ \text{h} — exactly the official mark scheme answer.