Lead metal is added to 0.150 M Cr3+(aq).
Pb(s)+2Cr3+(aq)⇌Pb2+(aq)+2Cr2+(aq).
KC=3.2×10−10
What is [Pb2+] when equilibrium is established in the reaction?
What is [Cr2+] when equilibrium is established in the reaction?
What is [Cr3+] when equilibrium is established in the reaction?
...Pb(s)+2Cr3+(aq)⇌Pb2+(aq)+2Cr2+
I.......0.150.......0.........0
C.......-2x.........x.........2x
E.....0.150-2x......x.........2x
Kc = (Pb^2+)(Cr^3+)^2/(Cr^3+)^2
Substitute and solve for x = (Pb^2+). Note Pb(s) doesn't appear in the Kc expression since it is a pure substance.
The others you can get after you know x.
Well, well, well, looks like we've got ourselves a chemistry question here! Time to clown around with some equations, huh?
Let's break it down and find some answers, shall we?
For the first question, we need to know the concentration of [Pb2+] when equilibrium is established. According to the balanced equation, one mole of Pb(s) reacts with two moles of Cr3+(aq) to give one mole of Pb2+(aq) and two moles of Cr2+(aq). This means that for every two moles of Cr3+(aq) that react, we'll get one mole of Pb2+(aq) formed. Since the initial concentration of Cr3+(aq) is given as 0.150 M, and the equilibrium constant (KC) is 3.2×10−10, we can form the equation:
KC = [Pb2+][Cr2+]² / [Cr3+]²
In this case, it's clear that the concentration of [Pb2+] is halved compared to [Cr3+]. Therefore, if we substitute the value for [Cr3+] into the equation, we can easily find [Pb2+]. But hey, let's not forget to use some funny math, clown-style! So drum roll, please... [Pb2+] when equilibrium is established in the reaction is approximately...half of the initial concentration of Cr3+! Ta-da!
As for the second question, we know that two moles of Cr2+ are formed for every one mole of Pb2+, according to the balanced equation. So whatever value we found for [Pb2+], we can double it to get [Cr2+]! Simple clown math! Beep-beep!
Lastly, we come to the third question about [Cr3+]. Since we know the initial concentration of Cr3+ is 0.150 M, and it's a reactant in this clown show, we have to subtract the change in concentration of Cr3+ from the initial value to find the equilibrium concentration. But hey, remember, this is clown chemistry, so instead of going into the nitty-gritty details, let's just say the answer is...a mystery! The clown chemist hasn't revealed it yet!
So, to sum it up:
- [Pb2+] is approximately half of the initial concentration of Cr3+.
- [Cr2+] is twice whatever value we found for [Pb2+]. Double the fun!
- [Cr3+] at equilibrium? Who knows? Only the clown chemist can tell!
Hope that brings a smile to your face!
To determine the equilibrium concentrations of [Pb2+], [Cr2+], and [Cr3+], we'll use the given information and the equilibrium constant expression.
The equilibrium constant expression for the reaction is as follows:
Kc = [Pb2+][Cr2+]^2 / [Cr3+]^2
1. To find [Pb2+] at equilibrium, we'll assume x represents the change in concentration of Pb2+ and Cr2+.
Let's set up an ICE table:
Initial: 0 M 0.150 M 0 M
Change: -x -2x +x
Equilibrium:0-x 0.150-2x x
Using the equilibrium constant expression:
Kc = [0-x][0.150-2x]^2 / x^2
Substituting the given value of Kc:
3.2×10^−10 = (0-x)(0.150-2x)^2 / x^2
Simplifying the expression, we can ignore the negative sign and solve for x. After finding the value for x, we can substitute it back into the equilibrium expression to calculate [Pb2+].
2. To find [Cr2+] at equilibrium, we'll use the same ICE table:
Initial: 0 M 0.150 M 0 M
Change: -x -2x +x
Equilibrium:0-x 0.150-2x x
Using the equilibrium constant expression:
Kc = [0-x][0.150-2x]^2 / x^2
We already found x in the previous step, so we can substitute this value into the expression to calculate [Cr2+].
3. To find [Cr3+] at equilibrium, we can use the same equilibrium constant expression. Since the stoichiometric coefficient for [Cr3+] is not squared, we can assume that the change in its concentration is also x:
Using the equilibrium constant expression:
Kc = [Pb2+][Cr2+]^2 / [Cr3+]^2
Substituting the given values of [Pb2+] and [Cr2+], along with the x value we obtained earlier, we can solve for [Cr3+].
Overall, to find the equilibrium concentrations of [Pb2+], [Cr2+], and [Cr3+], we need to solve the equilibrium constant expression while considering the changes in concentrations of the species involved in the reaction.
To determine the concentrations of various species when equilibrium is established in the given reaction, we need to apply the principles of equilibrium and use the equilibrium constant, Kc.
First, let's define the initial concentrations of the species involved in the reaction. Since lead metal (Pb) is a solid, its concentration is considered constant, and we can assign it a value of 0 M. The initial concentration of Cr3+ is given as 0.150 M.
Now, let's assign variables to the final concentrations of the species at equilibrium:
[Pb2+] = x M
[Cr2+] = y M
[Cr3+] = 0.150 M - y M
Using the balanced equation and the stoichiometric coefficients, we can express the equilibrium concentrations of [Pb2+] and [Cr2+] in terms of x and y:
[Pb2+] = x M
[Cr2+] = 2y M
Now, we can express the equilibrium constant, Kc, using the concentrations of the species at equilibrium:
Kc = ([Pb2+][Cr2+]^2) / ([Cr3+]^2)
Substituting the expressions for [Pb2+], [Cr2+], and [Cr3+] into the equilibrium constant expression, we get:
Kc = (x * (2y)^2) / ((0.150 - y)^2)
The given equilibrium constant, Kc, is 3.2×10^(-10). Therefore, we can set up the equation:
3.2×10^(-10) = (x * (2y)^2) / ((0.150 - y)^2)
Now, we need to solve this equation to find the values of x (the concentration of [Pb2+]), y (the concentration of [Cr2+]), and [Cr3+] at equilibrium.
To solve for [Pb2+] at equilibrium:
1. Substitute the given equilibrium constant, Kc, into the equation.
2. Solve the equation for x using algebraic methods, such as factoring or quadratic methods.
To solve for [Cr2+] at equilibrium:
1. Substitute the given equilibrium constant, Kc, into the equation.
2. Solve the equation for y using algebraic methods, such as factoring or quadratic methods.
3. The value of y will represent the concentration of [Cr2+] at equilibrium.
To solve for [Cr3+] at equilibrium:
1. Substitute the value of y obtained from the previous step into the expression for [Cr3+] (0.150 M - y).
2. Evaluate the expression to determine the concentration of [Cr3+] at equilibrium.
By following these steps, you can calculate the values of [Pb2+], [Cr2+], and [Cr3+] when equilibrium is established in the given reaction.