A mixture, containing 0.0750 M HCl(g) and 0.0330 M O2(g) is allowed to come to equilibrium at 480degreeC.
4HCl(g) + O2(g) --> 2Cl2(g) + 2H2O(g)
At equilibrium [Cl2] = 0.030 M. What is the value of Kc?
To find the value of Kc for the given reaction, we need to write the equilibrium expression and calculate the concentrations of the reactants at equilibrium.
The equilibrium expression for the reaction is:
Kc = [Cl2]^2/[HCl]^4[O2]
First, we need to calculate the concentrations of HCl and O2 at equilibrium. We can use the stoichiometry of the reaction to determine the changes in the concentrations.
From the balanced equation, we can see that 4 moles of HCl react with 1 mole of O2 to produce 2 moles of Cl2 and 2 moles of H2O. Therefore, the moles of Cl2 formed will be twice the moles of HCl reacted.
Let's assume x moles of HCl reacted. Then, 2x moles of Cl2 will be formed. Since the initial concentration of HCl is 0.0750 M, the concentration of HCl at equilibrium will be (0.0750 - x) M.
Using stoichiometry, we can also determine that the number of moles of O2 reacted will be x/4. Therefore, the concentration of O2 at equilibrium will be (0.0330 - x/4) M.
Given that [Cl2] at equilibrium is 0.030 M, we can substitute this value into the equation:
0.030 = (2x)^2/[(0.0750 - x)^4(0.0330 - x/4)]
Now, we can solve this equation to find the value of x, which represents the moles of HCl reacted.
However, solving this equation directly is complicated due to the high power term in the denominator. To simplify the equation, we can make some assumptions:
1. Because [Cl2] is given as 0.030 M, we can assume that x is small in comparison to 0.0750. Therefore, we can approximate (0.0750 - x) as 0.0750.
2. Similarly, we can approximate (0.0330 - x/4) as 0.0330.
Under these assumptions, the equation becomes:
0.030 = (2x)^2/(0.0750^4 * 0.0330)
Now, we can solve this equation:
0.030 = 4x^2/(0.0750^4 * 0.0330)
Rearranging the equation, we get:
x^2 = (0.030 * 0.0750^4 * 0.0330)/4
x^2 = 0.24912075
Taking the square root of both sides:
x = 0.499120711
Since x represents the moles of HCl reacted, we can substitute this value back into the equation to find the concentrations of HCl and O2 at equilibrium:
[Cl2] = 2x = 0.998241422 M
[HCl] = 0.0750 - x = 0.0750 - 0.499120711 = 0.575879289 M
[O2] = 0.0330 - x/4 = 0.0330 - 0.499120711/4 = 0.207269822 M
Now, we can substitute these values into the equilibrium expression to calculate the value of Kc:
Kc = [Cl2]^2/[HCl]^4[O2] = (0.998241422)^2 / (0.575879289)^4 * (0.207269822) = 0.6989
Therefore, the value of Kc for the given reaction is approximately 0.6989.
To find the value of Kc, we need to set up the expression for the equilibrium constant based on the balanced chemical equation:
4HCl(g) + O2(g) ⇌ 2Cl2(g) + 2H2O(g)
The expression for the equilibrium constant can be written as:
Kc = ([Cl2]²/[HCl]⁴[O2])
Given that [Cl2] at equilibrium is 0.030 M, we can substitute this value into the equation:
Kc = (0.030)²/([HCl]⁴[O2])
Now we need to find the concentrations of [HCl] and [O2].
Given that the initial concentration of HCl is 0.0750 M and the initial concentration of O2 is 0.0330 M, we need to determine whether the concentrations have changed at equilibrium.
Since the reaction involves the consumption of HCl and the formation of Cl2, the concentration of Cl2 at equilibrium is 0.030 M, we know that the concentration of HCl has decreased.
However, we need more information to determine the concentration of O2 at equilibrium. Is O2 a limiting reactant or in excess? To answer this, we need to compare the stoichiometric ratio of HCl to O2 in the balanced chemical equation.
The stoichiometric ratio of HCl to O2 is 4:1. Given that the initial concentration of HCl is 0.0750 M, if O2 is in excess, then its concentration would not have changed, and we would use the initial concentration of 0.0330 M. But if O2 is the limiting reactant, we need to calculate its concentration at equilibrium.
To determine whether O2 is the limiting reactant, we calculate the number of moles of each reactant:
moles of HCl = (initial concentration of HCl) x (volume of mixture)
= (0.0750 M) x (volume of mixture)
moles of O2 = (initial concentration of O2) x (volume of mixture)
= (0.0330 M) x (volume of mixture)
Now we compare the stoichiometric ratio of moles of HCl to O2:
moles of HCl : moles of O2 = 4 : 1
If the ratio is less than 4:1, O2 is the limiting reactant, and if the ratio is greater than 4:1, HCl is the limiting reactant.
Once we have determined whether O2 is the limiting reactant or not, we can calculate its concentration at equilibrium.
If O2 is in excess, we use the initial concentration of 0.0330 M. If O2 is the limiting reactant, we use the moles of O2 calculated above to determine its concentration at equilibrium.
Substituting the values of [Cl2], [HCl], and [O2] into the equation for the equilibrium constant:
Kc = (0.030)²/([HCl]⁴[O2])