2. When 1.00 mole of NH3 gas and 0.40 moles of N2 gas are placed in a 5.0 L container and allowed to reach an equilibrium at a certain temperature, it is found that 0.78 moles of NH3 is present. The reaction is: 2 NH3(g) <=> 3 H2(g) + N2(g)
Find Keq
To find the equilibrium constant (Kₑq) for the given reaction, we need to first write the balanced equation:
2 NH3(g) <=> 3 H2(g) + N2(g)
Next, we can construct an ICE table to determine the initial, change, and equilibrium concentrations of the reactants and products:
Initial:
NH3(g): 1.00 mol
N2(g): 0.40 mol
H2(g): 0 mol (not present initially)
Change:
NH3(g): -0.78 mol (decreases by this amount because 0.78 mol is consumed)
N2(g): -0.13 mol (decreases by this amount because the coefficient ratio is 2:1)
H2(g): +0.39 mol (increases by this amount because the coefficient ratio is 2:3)
Equilibrium:
NH3(g): 1.00 mol - 0.78 mol = 0.22 mol
N2(g): 0.40 mol - 0.13 mol = 0.27 mol
H2(g): 0 mol + 0.39 mol = 0.39 mol
Now, we can use the equilibrium concentrations to calculate Kₑq.
Kₑq = [H2]³ * [N2] / [NH3]²
Plugging in the values:
Kₑq = (0.39 mol)³ * (0.27 mol) / (0.22 mol)²
Kₑq = 0.0853
Therefore, the equilibrium constant (Kₑq) for this reaction is approximately 0.0853.