In a 1.00 L container at 272 degrees C 1.3 mol of N2 and 1.65 mol of H2 are added at equilibrium 0.100 mol NH3 is present calculate the equilibrium concentrations of N2 and H2
2NH3(g)<-->N2(g)+3H2(g)
Have you tried to read what you post with no periods? Where the hell does the sentence end and the next one start. Fortunately I know enough chemistry that I may be able to guess. You can hope that I guess right.
........2NH3 ==> N2 + 3H2
I........0......1.3....1.65
C.......+2x......-x....-3x
E....... 2x....1.3-x...1.65-3x
The problem tells you that 2x = 0.100.
Solve for x and evaluate 1.3-x and 1.65-3x
To calculate the equilibrium concentrations of N2 and H2, we need to use the stoichiometry of the reaction and the concept of the equilibrium constant, K.
The equilibrium constant expression for the given reaction is:
K = [N2][H2]^3 / [NH3]^2
First, let's calculate the initial concentrations of N2 and H2 before the reaction.
Initial concentration of N2 (in mol/L):
[N2] = (moles of N2) / (volume of container in L) = 1.3 mol / 1.00 L = 1.3 M
Initial concentration of H2 (in mol/L):
[H2] = (moles of H2) / (volume of container in L) = 1.65 mol / 1.00 L = 1.65 M
Now, let's assume the change in concentration of NH3 is "x" (in mol/L).
Change in concentration of N2:
[N2] = 1.3 M
Change in concentration of H2:
[H2] = 1.65 M
Change in concentration of NH3:
[NH3] = 0.100 M + x
According to the stoichiometry of the reaction, the change in concentration of N2 is -x (as the coefficient of N2 is 1), and the change in concentration of H2 is -3x (as the coefficient of H2 is 3).
So, at equilibrium, the concentrations will be:
[N2] = 1.3 M - x
[H2] = 1.65 M - 3x
[NH3] = 0.100 M + x
Now, substitute the equilibrium concentrations into the equilibrium constant expression and solve for x:
K = [N2][H2]^3 / [NH3]^2
K = (1.3 - x)(1.65 - 3x)^3 / (0.100 + x)^2
Now, you can solve this equation to find the value of x. Once you determine the value of x, substitute it back into the equilibrium expressions for [N2] and [H2] to obtain their equilibrium concentrations.
To calculate the equilibrium concentrations of N2 and H2, we can use the information provided in the problem along with the balanced chemical equation.
First, let's set up a table to keep track of the initial amounts, changes, and equilibrium concentrations of the substances involved:
Substance | Initial (mol) | Change (mol) | Equilibrium (mol)
N2 | 1.3 | ? |
H2 | 1.65 | ? |
NH3 | 0.1 | ? |
Next, we need to determine the changes in the number of moles for each substance. According to the balanced chemical equation, 1 mol of N2 reacts with 3 mol of H2 to produce 2 mol of NH3.
Since the initial amount of NH3 is 0.1 mol, this implies that 0.05 mol of N2 reacted (2 mol of NH3 is formed for every 1 mol of N2). Likewise, 0.15 mol of H2 reacted (3 mol of H2 is needed for every 1 mol of N2). Therefore, the changes in the number of moles are:
Substance | Initial (mol) | Change (mol) | Equilibrium (mol)
N2 | 1.3 | -0.05 |
H2 | 1.65 | -0.15 |
NH3 | 0.1 | +0.1 |
To find the equilibrium concentrations, we simply add the changes to the initial amounts:
Substance | Initial (mol) | Change (mol) | Equilibrium (mol)
N2 | 1.3 | -0.05 | 1.3 - 0.05 = 1.25
H2 | 1.65 | -0.15 | 1.65 - 0.15 = 1.50
NH3 | 0.1 | +0.1 | 0.1 + 0.1 = 0.2
Therefore, the equilibrium concentrations of N2, H2, and NH3 are 1.25 mol/L, 1.50 mol/L, and 0.2 mol/L, respectively.