Answer:
28,359.1 J/mol
Step-by-step explanation:
ΔG = ΔG° + RTln(Q)
the given conditions are non-standard, so calculate the values of Q and ΔG°.
the reaction for the dissociation of nitrous acid is:
HNO₂ ⇌ H⁺ + NO₂⁻
the equilibrium constant expression for the reaction is:
K = [H⁺][NO₂⁻]/[HNO₂]
At pH = 1.30, [H⁺] = 10^(-pH) = 5.01×10^(-2) M
substitute the given values into the expression:
K = (5.01×10^(-2))(6.0×10^(-4))/(0.25) = 1.203×10^(-5)
To calculate ΔG°, use the relationship between ΔG° and K:
ΔG° = -RTln(K)
Substitutethe values of R, T, and K:
ΔG° = -(8.314 J/mol·K)(298 K)ln(1.203×10^(-5)) = 27,615.7 J/mol
use the equation for ΔG:
ΔG = ΔG° + RTln(Q)
Substitute the values of ΔG°, R, T, and Q:
ΔG = 27,615.7 J/mol + (8.314 J/mol·K)(298 K)ln((5.01×10^(-2))(6.0×10^(-4))/(0.25)) = 28,359.1 J/mol
Therefore, the value of ΔG for this solution of nitrous acid is 28,359.1 J/mol.