A) For the reaction CH 4 (g) + H 2 O(g) ---> 3 H 2 (g) + CO(g) delta H° = 206.1

Question

A) For the reaction
CH4(g) + H2O(g) ---> 3H2(g) + CO(g)
delta H° = 206.1 kJ and delta S° = 214.7 J/K
The equilibrium constant for this reaction at 325.0 K is _____
Assume that delta H° and delta S° are independent of temperature.

B) Consider the reaction

2N2(g) + O2(g) ---> 2N2O(g)
Using the standard thermodynamic data in the tables linked above, calculate deltaGrxn for this reaction at 298.15K if the pressure of each gas is 35.48 mm Hg.
______ kJ/mol

C) Consider the reaction
CH4(g) + H2O(g) ---> 3H2(g) + CO(g)
Using the standard thermodynamic data in the tables linked above, calculate deltaG for this reaction at 298.15K if the pressure of each gas is 12.05 mm Hg.

______ kJ/mol

D) Consider the reaction

2N2(g) + O2(g) -----> 2N2O(g)
Use the standard thermodynamic data in the tables linked above. Calculate deltaG for this reaction at 298.15K if the pressure of N2O(g) is reduced to 10.45 mm Hg, while the pressures of N2(g) and O2(g)remain at 1 atm.
______ kJ/mol


Please check your calculations and 100% be correct

Explanation / Answer

A)
G = H - T*S
= 206100 - 325.0*214.7
= 136322.5 J

Now use:
G = -R*T*ln K
136322.5 = -8.314*325*ln K
K = 1.23*10^-22
Answer: 1.23*10^-22

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