Question

Calculate the equilibrium constant at 48 K for a reaction with ΔH_rxn^o = 10 kJ and ΔS_rxn^o = -100 J/K. (Don't round unil the end. Using the exponent enlarges any round-off error.)

Calculate the equilibrium constant at 102 K for the thermodynamic data in the previous question.

Answer 1

Given that , temperature = 48 K

ΔH_rxn^o = 10 kJ and ΔS_rxn^o = -100 J/K.

Now,use the ΔHo and ΔSo given to calculate ΔGo as follows:

ΔGo = ΔHo - TΔSo

= 10*1000 – 48 * ( - 100)

= 10,000+4800

= 14800 J/ mole

= 14.8 KJ/ mole

Then, we can calculate the equilibrium constant using the equation:

ΔGo = - RT ln Keq

Or

Keq = e^ - ΔGo / RT

= e^ - 14800 /8.3145* 48

= e^ -37.1

= 7.72*10^-17

Given that , temperature = 102 K

ΔH_rxn^o = 10 kJ and ΔS_rxn^o = -100 J/K.

Now,use the ΔHo and ΔSo given to calculate ΔGo as follows:

ΔGo = ΔHo - TΔSo

= 10*1000 – 102 * ( - 100)

= 10,000+10200

= 20200 J/ mole

= 20.2 KJ/ mole

Then, we can calculate the equilibrium constant using the equation:

ΔGo = - RT ln Keq

Or

Keq = e^ - ΔGo / RT

= e^ - 20200 /8.3145* 102

= e^ -23.82

= 4.52*10^-11