Question

Answer the following questions using the chemical reaction and thermochemical information given below:

NH3(g) + HI(g) ⇌ NH4I(s)

ΔHf° (kJ/mol) S° (J mol-1 K-1)

NH4I -201.40 186.90

NH3 -45.94 192.77

HI 26.36 206.59

1. Determine ΔG°rx (in kJ) for this reaction at 567.27 K. Assume ΔH°f and S° do not vary as a function of temperature. Report your answer to two decimal places

2. Determine the equilibrium constant for this reaction at 567.27 K. Report your answer to three significant figures in scientific notation.

3. If the partial pressure of HI is 7.87 atm and the partial pressure of NH3 is 5.81 atm, determine ΔG (in kJ) for this reaction at 567.27 K. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ)

Answer 1

1. Determine ΔG°rx (in kJ) for this reaction at 567.27 K. Assume ΔH°f and S° do not vary as a function of temperature. Report your answer to two decimal places

dG°rx = dH°rx - T dS°rx

First calculate the dH°rx and dS°rx for this reaction:

dH°rx = dH product – dH reactants

dH°rx = -201.40 kJ/mol – (-45.94 kJ/mol+ 26.36 kJ/mol)

dH°rx = -181.82 kJ/mol

dS°rx = dS product – dS reactants

dS°rx = 186.90 J/mol – (192.77 J/mol+ 206.59 J/mol)

dS°rx = -212.46 J/mol

dS°rx = - 0.2125 K J/mol

Now use this reaction:

dG°rx = dH°rx - T dS°rx

T= 567.27 K

dG°rx = -181.82 kJ/mol - 567.27 K* - 0.2125 K J/mol

dG°rx = -61.3 kJ/mol

2. Determine the equilibrium constant for this reaction at 567.27 K. Report your answer to three significant figures in scientific notation.

ln K eq = - dG⁰ / RT

or

K eq = e^- dG⁰ / RT

K eq = e^- dG⁰ / RT

Here, dG⁰= -61.3 kJ/mol

R = 0.008314 kJ mol^-1 K^-1. , T=567.27 K

K eq = e^ - (-61.3 kJ/mol ) / 0.008314 kJ mol^-1 K^-1. * 567.27 K

K eq = e^ 61.3 kJ/mol / 4.72kJ mol^-1

K eq = e^ 12.98

K eq =4.34*10^5

3. If the partial pressure of HI is 7.87 atm and the partial pressure of NH3 is 5.81 atm, determine ΔG (in kJ) for this reaction at 567.27 K. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ)

To calculate the dG at 567.27 K use the following expression

dG = dG^o + R T ln Q.

dG = -61.3 kJ/mol + 0.008314 kJ mol^-1 K^-1* 567.27 K ln 1/ 5.81 *7.87

dG = -61.3 kJ/mol + 0.008314 kJ mol^-1 K^-1* 567.27 K ln 1/ 45.72

dG = -61.3 kJ/mol + 0.008314 kJ mol^-1 K^-1* 567.27 K *(3.822)

dG=-61.3 kJ/mol - 18.028 kJ/mol

dG = 43.27 kJ/mol