Answer:
[tex]\large \boxed{\Delta G = \text{60 kJ}}[/tex]
Explanation:
Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻; Ksp = 1.12× 10⁻¹²
I/mol·L⁻¹: 0.191 0.823
To get ΔG under these conditions, we can use the equation
ΔG = ΔG° + RTlnQ = -RTlnK + RTlnQ = RTln[Q/K]
1. Calculate Q
[tex]Q =\text{[Ag$^{+}$]$^{2}$[Cr$_{2}$O$_{4}^{2-}$]} = 0.191^{2}\times0.823 = 0.03002[/tex]
2. Calculate ΔG
T = (25.0 + 273.15) K = 298.15 K
[tex]\begin{array}{rcl}\Delta G& =& RT \ln \left (\dfrac{Q}{K} \right )\\\\&=& 8.314 \times 298.15 \times \ln\left (\dfrac{0.03002}{1.12 \times 10^{-12}} \right )\\\\&=& 2479 \times \ln(2.860 \times10^{10})\\&=& 2479 \times 24.01\\& = & \text{59520 J}\\& = & \textbf{60 kJ} \\\end{array}\\\large \boxed{\mathbf{\Delta G} = \textbf{60 kJ}}[/tex]
