Question

The equilibrium constant for the reaction Ni2+(aq) + 6 NH3(aq) ⇌ Ni(NH3)6 2+(aq) is Kf = 5.6 × 108 at 25°C. (a) What is ΔG o at this temperature? (b) If standard-state concentrations of reactants and products are mixed, in which direction does the reaction proceed? (c) Determine ΔG when [Ni(NH3)62+] = 0.010 M, [Ni2+] = 0.0010 M, and [NH3] = 0.0050 M. In which direction will the reaction proceed to achieve equilibrium? (a) × 10 J/mol (Enter your answer in scientific notation.) (b) To the right. To the left. (c) × 10 J/mol (Enter your answer in scientific notation.) To the right. To the left.

Answer 1

Answer:

(a) -49.9 kJ/mol;

(b) To the right;

(c) 34.6 kJ/mol

Explanation:

(a) For this reaction, since it's at equilibrium and standard states, we know that we can apply the equation:

$\Delta G^o = -RT ln (K_f)$

Substituting the given variables:

$\Delta G^o = -8.314 \frac{J}{K mol}\cdot 298.15 K\cdot ln (5.6\cdot 10^8) = -49932 J/mol = -49.9 kJ/mol$

(b) Notice that this reaction is spontaneous, since $\Delta G^o < 0$. This means reaction spontaneously proceeds to the right side. Besides, K > 1, this means products dominate over reactants, so reaction proceeds to the right.

(c) Given the expression of the formation constant, we can use the same expression to calculate the reaction quotient at non-standard conditions:

$Q_f = \frac{[Ni(NH_3)_6]^{2+}}{[Ni^{2+}][NH_3]^6} = \frac{0.010}{0.0010\cdot 0.0050^6} = 6.4\cdot 10^{14}$

Now, notice that $Q_f > K_f$. In this case, we have an excess of the products, this means reaction will shift to the let left to restore the equilibrium.

Calculate:

$\Delta G = \Delta G^o + RT ln Q_f = -49932 J/mol + 8.314 \frac{J}{K mol}\cdot 298.15 K\cdot ln(6.4\cdot 10^{14}) = 34577 J = 34.6 kJ/mol$