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 2 of 5: Integrated concentration ratio
CstCs0=1K+1+KK+1exp⁡[−k−1t(K+1)]\frac{C_s^t}{C_s^0} = \frac{1}{K+1} + \frac{K}{K+1}\exp[-k_{-1}t(K+1)]
Analysis

Using k1=Kk−1k_1 = K k_{-1}, integrating ∫Cs0CstdCsCs−Cs0/(K+1)=−(K+1)k−1t\int_{C_s^0}^{C_s^t} \frac{\mathrm{d}C_s}{C_s - C_s^0/(K+1)} = -(K+1)k_{-1} t yields this standard form directly.