At room temperature (25 °C), the gas ClNO is impure because it decomposes slightly according to the equation
2ClNO(g) Cl2(g) + 2NO(g)
The extent of decomposition is about 4%. Based upon this, the approximate value of ΔG°298 for the reaction at that temperature is
I got my answer 25.4 kJ and 25.5 kJ but both was wrong. Did I do something wrong?
Did you use dGo formation for your values. I don't have those values. Were the two values separate calculations or is the question asking for two different values? I don't see but one question.
To calculate the approximate value of ΔG°298 for the reaction, you need to use the equation:
ΔG° = -RT ln(K)
Where:
- ΔG° is the standard Gibbs free energy change of the reaction.
- R is the gas constant (8.314 J/mol·K).
- T is the temperature in Kelvin (298 K).
- K is the equilibrium constant for the reaction.
First, let's find the equilibrium constant, K, using the extent of decomposition:
K = (concentration of products) / (concentration of reactants)
Since the extent of decomposition is given as 4%, we can assume that 4% of the initial ClNO decomposes to form Cl2 and NO, while the remaining 96% remains as ClNO.
Let x be the concentration of ClNO that decomposes, then the concentration of Cl2 and NO produced will be 2x.
So, the equilibrium constant becomes:
K = (2x)^2 / (0.96 - x)^2 ≈ (2x)^2 / (0.96)^2
Now, we need to convert the equilibrium constant to the natural logarithm form:
ln(K) = 2 ln(2x) - 2 ln(0.96)
Next, plug in the known values:
ln(K) = 2 ln(2(0.04)) - 2 ln(0.96)
ln(K) ≈ 2 ln(0.08) - 2 ln(0.96)
Now, calculate the value of ln(K) using a calculator:
ln(K) ≈ 2*(-2.5257) - 2*(-0.0408)
ln(K) ≈ -5.0514 + 0.0816
ln(K) ≈ -4.9698
Finally, substitute the values of R, T, and ln(K) back into the original equation to find ΔG°:
ΔG° = -RT ln(K)
ΔG° = -8.314 J/mol·K * 298 K * (-4.9698)
ΔG° ≈ 12400 J/mol
Converting the answer to kJ/mol:
ΔG° ≈ 12.4 kJ/mol
Therefore, the approximate value of ΔG°298 for the reaction at room temperature (25 °C or 298 K) is approximately 12.4 kJ/mol.