Question

Use the information in Table 5G.2 to determine the value of K at 500 K for the reaction 2NH3(g) + 3I2(g) ? N2(g) + 6HI(g).

Answer 1

Solution :-

Balanced reaction equation

2NH3(g) + 3I2(g) --- > N2(g) + 6HI(g)

Using the standard free energy of formation of the reactant and product we can find the free enrgy change of the reaction and then using the free energy change of the reaction we can find the equilibrium constant.

Delta G rxn = sum of delta Go product – sum of delta Go reactant

                      =[(N2*1)+(HI*6)]-[(NH3*2)+(I2*3)]

                      =[(0*1)+(1.71*6)]-[(-16.48*2)+(19.36*3)]

                       =-14.86 kJ

Now using the free energy value we can find the equilibrium constant

Delta G = - RT ln K

-14.86 kJ/mol = -( 0.008314 kJ/mol K)*500 K * ln K

-14.86 kJ/mol / -( 0.008314 kJ/mol K)*500 K = ln K

3.575 = ln K

K= e^(3.575)

K= 35.7

Therefore the equilibrium constant for the reaction is 35.7