Answer:
![\large \boxed{\Delta G = \text{60 kJ}}](https://img.qammunity.org/2021/formulas/chemistry/college/axlnk4e0ohl0kdo497cjul0ozmo7lp84xc.png)
Step-by-step 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
![Q =\text{[Ag$^(+)$]$^(2)$[Cr$_(2)$O$_(4)^(2-)$]} = 0.191^(2)*0.823 = 0.03002](https://img.qammunity.org/2021/formulas/chemistry/college/d5eb1wimnl1s8ym2o2to17pfqddtmkbl75.png)
2. Calculate ΔG
T = (25.0 + 273.15) K = 298.15 K
![\begin{array}{rcl}\Delta G& =& RT \ln \left ((Q)/(K) \right )\\\\&=& 8.314 * 298.15 * \ln\left ((0.03002)/(1.12 * 10^(-12)) \right )\\\\&=& 2479 * \ln(2.860 *10^(10))\\&=& 2479 * 24.01\\& = & \text{59520 J}\\& = & \textbf{60 kJ} \\\end{array}\\\large \boxed{\mathbf{\Delta G} = \textbf{60 kJ}}](https://img.qammunity.org/2021/formulas/chemistry/college/e1j21ehpy8hjflrn2886rx93d475i9m3oz.png)