A mixture consisting of 0.150 M N2(g) and 0.612 M H2(g) reacts and then reaches equilibrium according to the equation:
N2(g) + 3H2(g) ⇋ 2NH3(g)
At equilibrium, the concentration of ammonia is 0.213 M. Calculate the concentration (molarity) of H2(g) at equilibrium to 3 decimal places. Do NOT enter any units.
To find the concentration of H2(g) at equilibrium, we first need to understand the stoichiometry of the reaction and use the given information about the concentration of ammonia.
According to the balanced equation, the molar ratio between N2(g) and NH3(g) is 1:2, and the molar ratio between H2(g) and NH3(g) is 3:2.
Let x represent the change in the concentration of H2(g) at equilibrium.
At the start of the reaction, we have 0.150 M N2(g) and 0.612 M H2(g). Since the molar ratio between N2(g) and NH3(g) is 1:2, at equilibrium, the concentration of NH3(g) would be 2 * 0.150 = 0.300 M.
Using the given information, we know that at equilibrium, the concentration of NH3(g) is 0.213 M.
To find the change in the concentration of H2(g) at equilibrium, we can set up an ICE table:
Initial: 0.612
Change: -3x (since 3 H2(g) reacts to form 2 NH3(g))
Equilibrium: 0.612 - 3x
According to the molar ratio between H2(g) and NH3(g), the change in the concentration of H2(g) will be -3x.
Now, we need to use the equilibrium concentration of NH3(g) to solve for x:
0.213 = 2(0.612 - 3x)
0.213 = 1.224 - 6x
6x = 1.224 - 0.213
6x = 1.011
x = 1.011 / 6
x = 0.1685
Since we need the concentration of H2(g) at equilibrium, we can substitute the calculated value of x into the equation:
[H2(g)] = 0.612 - 3x
[H2(g)] = 0.612 - 3(0.1685)
[H2(g)] = 0.612 - 0.5055
[H2(g)] = 0.1065
Therefore, the concentration of H2(g) at equilibrium is 0.1065 M.
Note: It is always a good practice to double-check your answer and ensure the units are correct. In this case, we are not required to enter any units, but make sure to include the units when necessary in other calculations.