At a certain temperature, 0.885 mol of N2, and 2.654 mol of H2 are placed in a container.
N2+3H2=2NH3
At equilibrium, there is 0.823 mol of NH3 present. Determine the number of mol of N2 and H2 that are present when the reaction is at equilibrium.
Answer in moled of N2 at equilibrium and moles of H2 at equilibrium.
N2: 0.742 mol
H2: 2.412 mol
Well, it looks like our little N2 and H2 molecules are having a party in that container! But let's not forget about NH3 crashing the party too. Now, to figure out the number of moles of N2 and H2 at equilibrium, we can use the stoichiometry of the reaction.
From the balanced equation, we see that for every 1 mole of N2, we need 3 moles of H2 to produce 2 moles of NH3. So, let's do some mol-to-mol conversions to find the number of moles of N2 and H2 at equilibrium.
We start with 0.885 moles of N2. Since the ratio of N2 to NH3 is 1:2, we can say that for every mole of N2 reacted, 2 moles of NH3 are produced. So, let's multiply 0.885 moles of N2 by 2/1 to find the moles of NH3 produced: 0.885 * 2/1 = 1.77 moles of NH3 produced.
Now, we know that at equilibrium, there are 0.823 moles of NH3 present. So, if we subtract the moles of NH3 produced (1.77 moles) from the moles of NH3 present at equilibrium (0.823 moles), we can find the moles of NH3 that were consumed: 0.823 - 1.77 = -0.947 moles of NH3 consumed.
Now, since the ratio of NH3 to H2 is 2:3, we can say that for every 2 moles of NH3 reacted, 3 moles of H2 are consumed. So, let's multiply -0.947 moles of NH3 consumed by 3/2 to find the moles of H2 consumed: -0.947 * 3/2 = -1.4215 moles of H2 consumed.
Now, to find the moles of N2 and H2 at equilibrium, we subtract the moles consumed from the initial moles.
Moles of N2 at equilibrium = 0.885 - 0.823 = 0.062 moles
Moles of H2 at equilibrium = 2.654 - (-1.4215) = 4.0755 moles
So, the number of moles of N2 at equilibrium is approximately 0.062 moles, and the number of moles of H2 at equilibrium is approximately 4.076 moles.
Hope that clears things up! Enjoy the party in that container!
To determine the number of moles of N2 and H2 at equilibrium, we can use the stoichiometry of the balanced chemical equation.
Given:
Initial moles of N2 = 0.885 mol
Initial moles of H2 = 2.654 mol
Moles of NH3 at equilibrium = 0.823 mol
From the balanced chemical equation, we can see that the mole ratio of N2 to NH3 is 1:2, and the mole ratio of H2 to NH3 is 3:2.
1. Calculate the moles of N2 at equilibrium:
Moles of NH3 = 0.823 mol
Mole ratio of N2 to NH3 = 1:2
Moles of N2 = (Mole ratio of N2 to NH3) * (Moles of NH3)
= (1/2) * 0.823 mol
= 0.4115 mol
Therefore, at equilibrium, there are approximately 0.4115 mol of N2 present.
2. Calculate the moles of H2 at equilibrium:
Moles of NH3 = 0.823 mol
Mole ratio of H2 to NH3 = 3:2
Moles of H2 = (Mole ratio of H2 to NH3) * (Moles of NH3)
= (3/2) * 0.823 mol
= 1.2345 mol
Therefore, at equilibrium, there are approximately 1.2345 mol of H2 present.
To determine the number of moles of N2 and H2 present at equilibrium, we can use the balanced equation for the reaction:
N2 + 3H2 → 2NH3
First, let's calculate the initial moles of N2 and H2. Given that 0.885 mol of N2 and 2.654 mol of H2 were initially placed in the container, we have:
Initial moles of N2 = 0.885 mol
Initial moles of H2 = 2.654 mol
Next, let's determine the change in moles of N2 and H2. According to the balanced equation, for every 1 mole of N2 consumed, 3 moles of H2 are consumed, and 2 moles of NH3 are produced. Therefore, the stoichiometric ratio between N2 and H2 is 1:3. Since the reaction is at equilibrium and we know that 0.823 mol of NH3 is present, we can calculate the change in moles of N2 and H2:
Change in moles of N2 = 0.823 mol × (1 mol N2 / 2 mol NH3) = 0.412 moles of N2
Change in moles of H2 = 0.823 mol × (3 mol H2 / 2 mol NH3) = 1.235 moles of H2
Finally, to find the moles of N2 and H2 at equilibrium, we add the initial moles to the change in moles:
Moles of N2 at equilibrium = Initial moles of N2 + Change in moles of N2
= 0.885 mol + 0.412 mol
= 1.297 mol of N2
Moles of H2 at equilibrium = Initial moles of H2 + Change in moles of H2
= 2.654 mol + 1.235 mol
= 3.889 mol of H2
Therefore, at equilibrium, there are approximately 1.297 moles of N2 and 3.889 moles of H2 present.