Question

3. + 2.5/10 points Previous Answers McM8 6.P.012. The standard Gibbs-free energy of a system is related to its equilibrium constant through the following equation. AG° = -R·T· In(K) In this equation R is the gas constant, T is the temperature, and the ° next to AG defines the conditions as standard ambient temperature and pressure, i.e. "SATP". (Answer the following questions to three significant figures.) (a) Given an equilibrium constant of 4.53 x 10-6, what is its standard Gibbs-free energy? 49 30.5 kJ/mol (b) Given a standard Gibbs-free energy of 25.3 kJ/mol, what is the value of its equilibrium constant? Check the number of significant figures. (c) Suppose a reaction has a standard Gibbs-free energy of 4.87 kJ/mol, calculate the relative ratio of product to reactant. product reactant % 4.9 % More Information. Submit Answer Practice Another Version Viewing Saved Work Revert to Last Response

Answer 1

The relationship between standard gibbs free energy (ΔG^o) and equilibrium constant (Kc) is

ΔG^o = -RT lnK

where R = universal gas constant = 0.008314 kJ mol^-1 K^-1.

At standard condition, T = temperature = 298 K

Now putting these values-

1- Given Kc = 4.53 * 10^-6

Thus ΔG^o = -RT lnK

=  -0.008314 kJ mol^-1 K^-1. * 298 K*  ln(4.53 * 10^-6)  

= -0.008314 kJ mol^-1 K^-1. * 298 K*  (-12.3)  

= 30.47 kJ mol^-1

2- Given ΔG^o = 25.3 kJ mol^-1

Then ΔG^o = -RT lnK

lnK = -ΔG^o /RT

  = -25.3 kJ mol^-1 /0.008314 kJ mol^-1 K^-1. * 298 K

= -10.21

K = e^(-10.21)

= 3.68 * 10^-5

3-

Given ΔG^o = 4.87 kJ mol^-1

Then ΔG^o = -RT lnK

lnK = -ΔG^o /RT

  = -4.87 kJ mol^-1 /0.008314 kJ mol^-1 K^-1. * 298 K

= -1.965

K = e^(-1.965)

= 0.140

Again we know Equilibrium constant (Kc) is the ratio between the concentration of product to concentration of reactant at equilibrium

That means Kc = [product]/ [reactant] = 0.140

= 14 / 100

That means if at equilibrium, we have 100 unit of reactant present, then amount of product formed = 14 unit

Thus product : reactant = 14 : 100

= 7 : 50