Reaction A B proceeds with H = 240 kJ×mol-1; S = 40 J×K-1×mol-1. Substance A was taken at initial concentration 2 mM. Determine the concentrations of A and B after the reaction reaches equilibrium at 298 K.
To determine the concentrations of substances A and B at equilibrium, we can use the concept of Gibbs free energy (ΔG) and the equilibrium constant (K). The equilibrium constant (K) is calculated using the equation:
K = e^(-ΔG/RT)
Where:
- ΔG is the change in Gibbs free energy,
- R is the gas constant (8.314 J⋅K^(-1)⋅mol^(-1)),
- T is the temperature in Kelvin.
The change in Gibbs free energy (ΔG) can be calculated using the equation:
ΔG = ΔH - TΔS
Where:
- ΔH is the change in enthalpy,
- ΔS is the change in entropy.
Let's plug in the given values to calculate K and then determine the equilibrium concentrations.
Given:
ΔH = 240 kJ⋅mol^(-1) = 240,000 J⋅mol^(-1)
ΔS = 40 J⋅K^(-1)⋅mol^(-1)
T = 298 K
Initial concentration of A = 2 mM
First, convert the concentrations of A into mol⋅L^(-1). Since 1 mM = 1 mol⋅m^(-3), the initial concentration of A is 0.002 mol⋅L^(-1).
Now let's calculate ΔG:
ΔG = ΔH - TΔS = 240,000 J⋅mol^(-1) - (298 K)(40 J⋅K^(-1)⋅mol^(-1))
= 240,000 J⋅mol^(-1) - 11,920 J⋅mol^(-1)
= 228,080 J⋅mol^(-1)
Next, let's calculate K using the equation:
K = e^(-ΔG/RT) = e^(-228,080 J⋅mol^(-1) / (8.314 J⋅K^(-1)⋅mol^(-1))(298 K)
≈ e^(-96.72)
To determine the concentrations at equilibrium, we need the stoichiometry of the reaction. Without this information, we cannot determine the exact concentrations of A and B.
However, we can interpret the equilibrium constant (K) to determine some general information about the reaction. If K is greater than 1, it means the forward reaction is favored at equilibrium, and if K is less than 1, the reverse reaction is favored. If K is approximately equal to 1, it means the reaction is close to equilibrium.
I apologize for not being able to provide the exact concentrations of A and B without the stoichiometry. But we have determined the equilibrium constant (K) and its interpretation.