For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 5.00 L rigid container are [H2] = 0.0500 M, [F2] = 0.0100 M, and [HF] = 0.400 M. If 0.340 mol of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.
H2(g) + F2(g) 2 HF(g)
You have two problems here.
First, write Kc expression, substitute eq concns, and calculate Kc.
Then you can address the second half of the problem.Note:I have omitted all of th zeros to save space but you should include them.
..............H2+ F2 ==> 2HF
initial....0.05M.0.01M...0.400
add..............(0.340 mol/5L = ?)
change........-x..-x......+2x
eqil.....0.05-x..0.078-x...0.400+2x
Substitute the equil concns into the Kc expression and solve for x. Then plug x into the eq conditions of the ICE chart above and solve for the individual concns of each element/compound.
Oh, look! The gases are having a little party in that rigid container. But uh-oh, they're ruining the equilibrium! Let me grab my clown calculator and figure out what happens next.
To solve this problem, we can use the formula \( K_c = \frac{{[HF]^2}}{{[H_2][F_2]}} \), where \( K_c \) is the equilibrium constant.
Given that \([H_2] = 0.0500 \, \text{M}\), \([F_2] = 0.0100 \, \text{M}\), and \([HF] = 0.400 \, \text{M}\), we can substitute these values into the equilibrium constant formula to find the value of \( K_c \):
\( K_c = \frac{{(0.400)^2}}{{(0.0500)(0.0100)}} \)
Now, let's calculate this value of \( K_c \) and see what kind of party the equilibrium has in store for us!
Calculating... calculating... drumroll, please...
\( K_c = 3.20 \)
Alright, now we know the equilibrium constant. And since they gave us the change in concentration for \( F_2 \), we can assume that this process is happening under constant temperature and pressure. So, we can use the stoichiometry of the reaction to determine the changes in the concentrations of \( H_2 \), \( F_2 \), and \( HF \) due to the addition of \( 0.340 \, \text{mol} \) of \( F_2 \):
For every \( 1 \) mole of \( F_2 \) consumed, \( 2 \) moles of \( HF \) are formed. So, \( 0.340 \, \text{mol} \) of \( F_2 \) will consume \( 0.340 \, \text{mol} \times 2 = 0.680 \, \text{mol} \) of \( HF \).
Since \( F_2 \) is in excess, and the reaction goes in the forward direction to reestablish equilibrium, \( 0.680 \, \text{mol} \) of \( HF \) will be consumed, and \( 0.680 \, \text{mol} \div 2 = 0.340 \, \text{mol} \) of \( H_2 \) will be formed.
Now, we can subtract the changes from the initial concentrations to find the new concentrations at equilibrium:
[H2] = \( 0.0500 \, \text{M} + 0.340 \, \text{mol} \div 5.00 \, \text{L} \)
[F2] = \( 0.0100 \, \text{M} - 0.340 \, \text{mol} \div 5.00 \, \text{L} + 0.340 \, \text{mol} \div 5.00 \, \text{L} \)
[HF] = \( 0.400 \, \text{M} - 0.680 \, \text{mol} \div 5.00 \, \text{L} + 0.680 \, \text{mol} \div 5.00 \, \text{L} \)
Let me grab my clown calculator for one last time...
Calculating... crunching numbers... spinning plates...
[H2] = \( 0.116 \, \text{M} \)
[F2] = \( 0.040 \, \text{M} \)
[HF] = \( 0.120 \, \text{M} \)
Ta-da! The party's over, and the equilibrium is reestablished with the following concentrations: [H2] = \( 0.116 \, \text{M} \), [F2] = \( 0.040 \, \text{M} \), and [HF] = \( 0.120 \, \text{M} \). Hope you enjoyed my clownish explanation!
To calculate the concentrations of all gases once equilibrium is reestablished, we can use the concept of the equilibrium constant, Kc, which is determined by the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.
The equilibrium constant expression for the given reaction is:
Kc = [HF]² / ([H2] * [F2])
Given:
[H2] = 0.0500 M
[F2] = 0.0100 M
[HF] = 0.400 M
The equilibrium constant (Kc) at this temperature can be calculated by substituting the given concentrations into the equilibrium constant expression:
Kc = (0.400 M)² / (0.0500 M * 0.0100 M)
= 16.00 / 0.00500
= 3200
Since the reaction is already at equilibrium, the value of Kc is constant. Now, in order to find the new concentrations of gases when 0.340 mol of F2 is added, let's analyze the balanced equation and changes that occur:
H2(g) + F2(g) 2 HF(g)
From the balanced equation, we can see that for every 1 mole of F2 consumed, 2 moles of HF are produced. Therefore, since 0.340 mol of F2 is added, the moles of HF produced will be 0.340 mol * 2 = 0.680 mol.
To find new concentrations after adding F2, we need to calculate the total moles of each species and take into account the volume of the system which is given as 5.00 L.
[H2] = (moles of H2) / (volume of system)
[F2] = (moles of F2) / (volume of system)
[HF] = (moles of HF) / (volume of system)
The initial moles of each species can be calculated as follows:
[H2] = (0.0500 M) * (5.00 L) = 0.250 mol
[F2] = (0.0100 M) * (5.00 L) = 0.050 mol
[HF] = (0.400 M) * (5.00 L) = 2.000 mol
After adding 0.340 mol of F2, the total moles of each species will be:
[H2] = 0.250 mol (no change)
[F2] = 0.050 mol + 0.340 mol = 0.390 mol
[HF] = 2.000 mol + 0.680 mol = 2.680 mol
Lastly, to find the new concentrations:
[H2] = (0.250 mol) / (5.00 L) = 0.0500 M (no change)
[F2] = (0.390 mol) / (5.00 L) = 0.0780 M
[HF] = (2.680 mol) / (5.00 L) = 0.536 M
Therefore, once equilibrium is reestablished, the concentrations of all gases will be approximately:
[H2] = 0.0500 M (no change)
[F2] = 0.0780 M
[HF] = 0.536 M
To answer this question, we need to use the concept of equilibrium and the equilibrium expression for the given reaction. The equilibrium expression is derived from the balanced chemical equation and gives us the relationship between the concentrations of the reactants and products at equilibrium.
The equilibrium expression for the given reaction is:
Kc = ([HF]²) / ([H2] * [F2])
where Kc is the equilibrium constant.
Given that the initial concentrations are [H2] = 0.0500 M, [F2] = 0.0100 M, and [HF] = 0.400 M, we can substitute these values into the equilibrium expression to find the equilibrium constant (Kc):
Kc = (0.400)² / (0.0500 * 0.0100)
= 0.64 / 0.005
= 128
Now, we can use the equilibrium constant to calculate the concentrations of all gases once equilibrium is reestablished after adding 0.340 mol of F2 to the mixture.
Let's assume the new concentration of F2 is x, then the concentrations of H2 and HF can be expressed as (0.0500 - x) and (0.400 + 2x), respectively, using stoichiometry.
Now, we can use the equilibrium expression and the known value of Kc to solve for x:
Kc = ([HF]²) / ([H2] * [F2])
128 = [(0.400 + 2x)²] / [(0.0500 - x) * x]
Simplifying, we have:
128 = (0.16 + 0.8x + 4x²) / (0.0500x - x²)
Now, we can cross-multiply and rearrange the equation:
128(0.0500x - x²) = 0.16 + 0.8x + 4x²
Simplifying further:
6.4x - 128x² = 0.16 + 0.8x + 4x²
Rearranging the equation:
128x² + 0.4x - 0.16 = 0
Using the quadratic formula, we can solve for x:
x = (-0.4 ± √(0.4² - 4 * 128 * -0.16)) / (2 * 128)
After evaluating the equation, we will get two possible values for x. Let's assume the positive root for x.
Once we have the value for x, we can calculate the concentrations of all gases at equilibrium:
[F2] = 0.0100 M + x
[H2] = 0.0500 M - x
[HF] = 0.400 M + 2x
Substitute the value of x into these expressions to get the final concentrations of all gases.