Question

The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4. You may want to reference (Pages 833 - 836) Section19.7 while completing this problem.

1. What is the value of ΔG when [H+] = 5.9×10−2M , [NO−2] = 6.3×10−4M and [HNO2] = 0.19 M ?

Express your answer using three significant figure.

Answer 1

Acid dissociation constant, K = 4.5x10+ Temperature, T = 298 K Gas constant, R= 8.314 JK 'mol-? Standard free energy change of the reaction, AG = - RT in K =-(8.314 JK 'mol-1) (298 K)\n(4.5x10*) =-(8.314 JK 'mo1") (298 K)(-7.71) = +1.91*10* Ix 1kJ mol 10' J = +19.1 kJ/mol Equation : HNO, (aq) → H* (aq) + NO, (aq) Free energy change of the reaction AG = AGO+RT in ([H* [N0,-]) = AG° +RT IN THNO,] = { +19.1 kJ/mol ) + (8.314x103 kW*mo1")(298 10-)(6.3x10+) 0.19 = ( +19.1 kJ/mol ) + (8.314x10-3 kJ K+mol-)(298 K)(-8.54) =( +19.1 kJ/mol ) +(-21.16 kJ/mol) = -2.06 kJ/mol = -2.10 kJ/mol