Chemical Thermodynamics - SS3 Chemistry Past Questions and Answers - page 3

The relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K) for a reaction at a given temperature (T) is given by the equation:

ΔG° = -RT ln(K)

where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin.

Given ΔG° = -20.0 kJ/mol and T = 298 K, we need to convert ΔG° to J/mol:

ΔG° = -20.0 kJ/mol × 1000 J/1 kJ = -20,000 J/mol

Now, we can calculate the equilibrium constant (K):

K = e(-ΔG° / (RT))

K = e^(-(-20,000 J/mol) / (8.314 J/(mol·K) × 298 K))

K = e^(67.94)

K ≈ 9.42 × 10²⁹

Therefore, the equilibrium constant (K) for the reaction at 298 K is approximately 9.42 × 10²⁹.

The reaction quotient (Q) is calculated using the same formula as the equilibrium constant (K), but instead of using standard Gibbs free energy change (ΔG°), we use the concentrations of the reactants and products at a given moment. The formula is:

Q = [NH3]² / ([N2] × [H2]³)

where [N2], [H2], and [NH3] are the concentrations of N2, H2, and NH3, respectively.

Given [N2] = 0.10 M, [H2] = 0.20 M, and [NH3] = 0.30 M, we can calculate the value of Q:

Q = (0.30 M)² / (0.10 M × (0.20 M)³)

Q = 0.09 M² / (0.10 M × 0.008 M³)

Q = 0.09 M² / (0.0008 M⁴)

Q ≈ 112.5 M^(-2)

Therefore, the value of the reaction quotient (Q) for this system is approximately 112.5 M^(-2).

Hess's Law is a fundamental principle in thermodynamics that states the enthalpy change of a reaction is independent of the reaction pathway taken from reactants to products. In other words, the overall enthalpy change of a reaction remains the same, regardless of the intermediate steps involved in the reaction.

To calculate the enthalpy change for the given reaction, we can use Hess's Law. We need to manipulate the given equations to match the target reaction. Notice that the first equation is already the reverse of the target reaction. To reverse the second equation, we must change the sign of the enthalpy change:

N2(g) + 3H2(g) → 2NH3(g) ΔH = -92.4 kJ/mol

2NO(g) → N2(g) + O2(g) ΔH = -180.6 kJ/mol (reversed)

Now, we can add the manipulated equations to obtain the target reaction:

N2(g) + 3H2(g) → 2NH3(g) ΔH = -92.4 kJ/mol

2NO(g) → N2(g) + O2(g) ΔH = -180.6 kJ/mol (reversed)

Target: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g) ΔH = -273.0 kJ/mol

The enthalpy change for the target reaction is -273.0 kJ/mol. However, since the enthalpy change is for the formation of the products, we change the sign to make it positive:

Enthalpy change for the target reaction = -(-273.0 kJ/mol) = 273.0 kJ/mol

Therefore, the correct answer is A) -91.8 kJ/mol.