Answer:
The value of ΔG of the reaction at 298 K is -6.495 kJ/mol.
Explanation:
[tex]P_2(g)+3Cl_2(g)\rightarrow 2PCl_3(g)[/tex]
partial pressure of phosphorus =[tex]=p_{P_2}=1.5 atm[/tex]
partial pressure of chlorine gas= [tex]=p_{Cl_2}=1.6 atm[/tex]
partial pressure of phosphorus trichloride =[tex]=p_{PCl_3}=0.65 atm[/tex]
The expression of Quotient of the reaction:
[tex]Q=\frac{[p_{PCl_3}]^2}{[p_{P_2}][p_{Cl_2}]^3}[/tex]
[tex]Q=\frac{(0.65 atm)^2}{1.5 atm\times (1.6 atm)^3}=0.06877[/tex]
[tex]\Delta G=\Delta G^o+-RT\ln Q[/tex]
where,
ΔG = Gibbs's free energy at 298K
ΔG° = Gibbs's free energy at equilibrium = -642.9kJ/mol = -642900 J/mol
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = =298 K
[tex]Q[/tex] = Reaction quotient = 0.06877
Putting values in above equation, we get:
[tex]\Delta G=-642900 J/mol+(8.314J/Kmol)\times 298K\times \ln (0.06877)[/tex]
[tex]\Delta G=-649,532.43 J/mol=-6.495 kJ/mol [/tex]
The value of ΔG of the reaction at 298 K is -6.495 kJ/mol.
