The equilibrium constant for the reaction 2 C_3 H_6(g) C_2 H_4(g) + C_4 H_8 (g)

Question

The equilibrium constant for the reaction 2 C_3 H_6(g) C_2 H_4(g) + C_4 H_8 (g) Is found to fit the expression: Ink A + B/T + C/T^2, between 300 and 600 K, where A = -1.04. B = -1088 K, C = 1.51 times 10^5 K^2. Calculate the standard reaction enthalpy and standard reaction entropy at 400 K. Estimate the temperature at which CaCO_3 (calcite) decomposes. Calculate the standard Gibbs energy and the equilibrium constant at (a) 298 K and (b) 400 K, for the reaction: PbO(s) + CO(g) - Pb(s) + CO_2 (g) Assume that the enthalpy of the reaction is independent of temperature.

Explanation / Answer

1.

The equilibrium constant of the reaction

2C3H6 (g) <-------> C2H4 (g) + C4H8 (g)

Expression ln k= A + B/T + C/T2found to be fit between 300 K and 600 K.

Here A = - 1.04, B = – 1088K and C= 1. 51 x 10^ 5K^2

Step 1- calculate ln K at 390 K and 410 K

At 390 K-

ln K = 1. 04 1088/390 + 1. 51 x 10^ 5/(390)^2 = 2.83698

At 410 K’

ln K’ = 1.04 1088/410 + 1. 51 x 10^ 5/(410)^2 = 2.79538

Step 2- Calculate the standard reaction enthalpy-

The van’t Hoff equation yields-

ln k’ - ln K = Hr0/R (1/390 – 1/410)

Re-arranging for the enthalpy and substituting

Hr0 = R {2.79538 – ( 2.83698)}/ (1/390 – 1/410)

Hr0 = 8 .314Jmol.K (2.79538 +2.83698) /0.000125 = 2765 J = 2.765 KJ

Step 3- Calculate the standard reaction free energy change-

At 400 K,

ln K = 1. 04 – 1088 /400 + 1. 51 x 10^ 5/(400)^2 =–2.81625

Gr0 = RT lnK

           = (8.3145) (400)(-2.81625)

           = 9373 J/mol = 9.37 KJ/mol

Step 4-Calculate standard reaction entropy at 400 K

Sr0 = Hr0 -Gr0 /T

                     = 2765 J/mol -9370J/mol /400K

                    = 16 5. J K-1 mol -1

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