In a certain experiment, 1.000 mole of N2 (g) and 1.000 mole of H2 were allowed to react at 500°C in a 1.000-liter flask. The relevant reaction was as follows: N2 (g) + 3H2 (g) -> 2NH3 (g)
After the system reached equilibrium, the flask was found to contain 0.921 mole of N2 (g). Calculate the equilibrium concentrations of H2(g) and NH3(g).
Initial (N2) = 1 mo/L
Initial (H2) = 1 mol/L
........................N2 (g) + 3H2 (g) -> 2NH3 (g)
I.....................1.0............1.0................0
Change...........-x.............-3x..............2x
equil...............0.921..............................
change for N2 must be 1.000 - 0.921= 0.079 = x
Then 3x must be 3*0.079 = 0.237 and 1 - 3x will be ?
and 2x must be 2*0.079 = 0.158 and 0 + 2x will be ?
Post your work if you get stuck.
Okay, thank you! That makes sense now.
To calculate the equilibrium concentrations of H2(g) and NH3(g), first, we need to determine the amount of each substance that reacts and the amount that is formed at equilibrium.
Given:
Initial moles of N2(g) = 1.000 mole
Initial moles of H2(g) = 1.000 mole
The balanced equation shows that 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3. This means that, according to stoichiometry, 3 moles of H2 are needed to react with 1 mole of N2.
Since we have 0.921 moles of N2 remaining at equilibrium, we can calculate the moles of H2 that reacted:
(1.000 mole - 0.921 mole) = 0.079 mole of N2 reacted.
Using the stoichiometry, we can calculate the moles of H2 that reacted:
0.079 mole of N2 x (3 mole H2 / 1 mole N2) = 0.237 mole of H2 reacted.
Now, let's calculate the moles of NH3 formed at equilibrium:
0.079 mole of N2 x (2 mole NH3 / 1 mole N2) = 0.158 mole of NH3 formed.
To determine the equilibrium concentrations, we need to consider the total volume of the flask, which is 1.000 liter.
Concentration of H2(g) at equilibrium:
0.237 mole of H2 / 1.000 L = 0.237 M
Concentration of NH3(g) at equilibrium:
0.158 mole of NH3 / 1.000 L = 0.158 M
Therefore, the equilibrium concentrations of H2(g) and NH3(g) are 0.237 M and 0.158 M, respectively.
To calculate the equilibrium concentrations of H2(g) and NH3(g), we need to use the concept of the equilibrium constant (Kc) and the balanced equation of the reaction.
The balanced equation of the reaction is:
N2 (g) + 3H2 (g) -> 2NH3 (g)
The equilibrium constant expression for this reaction is:
Kc = [NH3(g)]^2 / [N2(g)] * [H2(g)]^3
Given that the initial moles of N2 (g) and H2 (g) were both 1.000 mole, and at equilibrium, there were 0.921 moles of N2 (g) remaining, we can calculate the change in moles of N2 (g) using the equation:
Change in moles of N2 (g) = Initial moles of N2 (g) - Equilibrium moles of N2 (g)
Change in moles of N2 (g) = 1.000 mole - 0.921 mole
Change in moles of N2 (g) = 0.079 mole
Since the stoichiometric coefficient of N2 (g) in the balanced equation is 1, the change in moles of NH3 (g) will be twice the change in moles of N2 (g):
Change in moles of NH3 (g) = 2 * Change in moles of N2 (g)
Change in moles of NH3 (g) = 2 * 0.079 mole
Change in moles of NH3 (g) = 0.158 mole
Since the stoichiometric coefficient of H2 (g) in the balanced equation is 3, the change in moles of H2 (g) will be three times the change in moles of N2 (g):
Change in moles of H2 (g) = 3 * Change in moles of N2 (g)
Change in moles of H2 (g) = 3 * 0.079 mole
Change in moles of H2 (g) = 0.237 mole
Now, we can calculate the equilibrium concentrations of H2(g) and NH3(g) using the given volume of the flask, which is 1.000 liter:
Concentration of H2(g) at equilibrium = Change in moles of H2(g) / Volume
Concentration of H2(g) = 0.237 mole / 1.000 liter
Concentration of H2(g) = 0.237 M (Molarity)
Concentration of NH3(g) at equilibrium = Change in moles of NH3(g) / Volume
Concentration of NH3(g) = 0.158 mole / 1.000 liter
Concentration of NH3(g) = 0.158 M (Molarity)
Therefore, the equilibrium concentration of H2(g) is 0.237 M, and the equilibrium concentration of NH3(g) is 0.158 M.