Question

Calculate the equilibrium constant for each of the reactions at 25 ∘C.

a. 2Cr3+(aq)+3Sn(s)→2Cr(s)+3Sn2+(aq)

b. O2(g)+2H2O(l)+2Sn2+(aq)→4OH−(aq)+2Sn4+(aq)

c. 2Cr3+(aq)+3Ni(s)→2Cr(s)+3Ni2+(aq)

Answer 1

The equilibrium constants for the reactions can be calculated using the standard free energy change and the formula K = e^{(-ΔG°/(RT))}, but specific ΔG° values or tables referenced are needed to complete the calculations.

Step-by-step explanation:

To calculate the equilibrium constant for the reactions at 25 °C, we use the given thermodynamic data from the tables and apply the relation between the standard free energy change (ΔG°) and the equilibrium constant (K). The relation is given by the equation ΔG° = -RTlnK, where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant. For each reaction with a given ΔG° value, we can rearrange the formula to solve for K: K = e^{(-ΔG°/(RT))}.

For the provided exercises:

1. For reaction (a) 2Cr3+(aq)+3Sn(s)→2Cr(s)+3Sn2+(aq), we would need the ΔG° values for each species to perform the calculation.
2. For reaction (b) O2(g)+2H2O(l)+2Sn2+(aq)→4OH-(aq)+2Sn4+(aq), this appears to be a similar reaction to the provided exercises, but without the exact ΔG° values or Table P2, the equilibrium constant cannot be calculated.
3. For reaction (c) 2Cr3+(aq)+3Ni(s)→2Cr(s)+3Ni2+(aq), again, the ΔG° values would be necessary to calculate K.

When the necessary thermodynamic data is provided, each reaction's equilibrium constant can be calculated using the formula K = e^{(-ΔG°/(RT))}, with R = 8.314 J/mol·K, and T = 298.15 K (since 25 °C is equivalent to 298.15 K).