Calculate Delta G given equilibrium constant and temperature PL2E DS& the values

Question

Calculate Delta G given equilibrium constant and temperature

PL2E DS& the values in the resources listed below instead of the textbook values. Calculate AG® (in k)/mol) for each of the following reactions from the equilibrium constant at the temperature given. (a) C00(s) +CO(g)=Co(s) + CO2(g) K= 218 T= 600.0°C kJ/mol (b) (CH3)3N(aq) + H2O(/) (CH3)3NH +(aq) + OH-(aq) T:: 25.0°C K= 6.4 x 10-5 kJ/mol 4 (c) N205(g)-N203(g) + O2(g) T= 600.0°C K= 1.6x10 kJ/mol (d) H2O(/)-H2O(g) T=5.0°C K=0.00876 kJ/mol kJ/mol kJ/mol Type here to search

Explanation / Answer

Solution:-

when equilibrium constant (Keq ) and the temperature is given then we can find out delta g by using the formula

delta G = - RT ln Keq ----------(1) Here R = 8.314 J mol-1K-1 and T = temperature in K

So by using this formula, we can find out delta G

A) T = 600 oC = 873.15K and K = 218 [ 0o C = 273.15 K ]

Put all the values in equation(1) delta G = - 8.314J mol-1K-1 * 873.15 k * ln(218) = -39088 J /mol = -39.088 KJ/mol

B) T = 25 oC = 298.15K and K = 6.4*10-5 [ 0o C = 273.15 K ]

delta G = - 8.314J mol-1K-1 * 298.15 k * ln(6.4*10-5 ) = 23937 J /mol = 23.937 KJ/mol

C) T = 600 oC = 873.15K and K = 1.6*104 [ 0o C = 273.15 K ]

delta G = - 8.314J mol-1K-1 * 873.15 K * ln(1.6*104 ) = -70273 J /mol = -70.273 KJ/mol

D) T = 5 oC = 278.15K and K = 0.00876 [ 0o C = 273.15 K ]

delta G = - 8.314J mol-1K-1 * 278.15 k * ln( 0.00876 ) = -10955.8 J /mol = -10.9558 KJ/mol

E) T = 530 oC = 803.15 K and K = 1.0*10-26 [ 0o C = 273.15 K ]

delta G = - 8.314J mol-1K-1 * 803.15 k * ln(1.0*10-25 ) = 399756.7 J /mol = 399.756 KJ/mol

F) T = 25 oC = 298.15K and K = 9.8*10-9 [ 0o C = 273.15 K ]

delta G = - 8.314J mol-1K-1 * 298.15 k * ln(9.8*10-9 ) = 45711.6 J /mol = 45.7116 KJ/mol

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