Consider the following reaction.
A(aq) (Two way arrows) 2B(aq)
Kc = 7.94*10^-6 at 500K
If a 2.70 M sample of A is heated to 500K what is the concentration of B at equilibrium?
Kc = ([x]^2)/A
7.94*10^-6 = ([x]^2)/2.70 M
2.1438*10^-5 = x^2
x = 4.630118789*10^-3
I hope this helps Janavi.
Thanks Karan!!
To find the concentration of B at equilibrium, we need to use the equilibrium constant (Kc) and the initial concentration of A.
Given:
Initial concentration of A (A0) = 2.70 M
Equilibrium constant (Kc) = 7.94 * 10^-6
Temperature (T) = 500K
The balanced equation for the reaction is:
A(aq) ⇌ 2B(aq)
Let x be the change in concentration of A and 2x be the change in concentration of B at equilibrium.
At equilibrium, the concentrations of A and B will be:
[A] = [A0] - x
[B] = 2x
According to the law of mass action, the expression for Kc is:
Kc = ([B]^2) / [A]
Kc = (2x)^2 / ([A0] - x)
Now, substitute the given values into the equation and solve for x.
7.94 * 10^-6 = (2x)^2 / (2.70 - x)
7.94 * 10^-6 = 4x^2 / (2.70 - x)
Cross multiply:
7.94 * 10^-6 * (2.70 - x) = 4x^2
Multiply and simplify:
2.1388 * 10^-5 - 7.94 * 10^-6x = 4x^2
Rearrange the equation to form a quadratic equation:
4x^2 + 7.94 * 10^-6x - 2.1388 * 10^-5 = 0
Solve this quadratic equation using the quadratic formula or a suitable method to find the value of x. Once you find the value of x, substitute it back into [B] = 2x to obtain the equilibrium concentration of B.
To find the concentration of B at equilibrium, we can use the equilibrium constant expression and set up an ICE table to calculate the changes in concentration. Here's how you can do it:
Step 1: Write the balanced equation for the reaction:
A(aq) ⇌ 2B(aq)
Step 2: Write the equilibrium constant expression:
Kc = [B]^2 / [A]
Step 3: Set up an ICE (Initial, Change, Equilibrium) table:
Let's assume the initial concentration of A is 2.70 M, and let's represent the change in concentration of B as 'x'.
A ⇌ 2B
Initial: 2.70 M 0 M
Change: -x M +2x M
Equilibrium: 2.70 - x M 2x M
Step 4: Substitute the equilibrium concentrations into the equilibrium constant expression:
Kc = (2x)^2 / (2.70 - x)
Step 5: Solve for 'x':
Kc = (2x)^2 / (2.70 - x)
Simplifying the equation:
7.94x10^-6 = 4x^2 / (2.70 - x)
Multiply both sides of the equation by (2.70 - x):
7.94x10^-6 * (2.70 - x) = 4x^2
Rearrange the equation:
4x^2 - 7.94x10^-6 * (2.70 - x) = 0
This is a quadratic equation. Solve for 'x' using the quadratic formula.
Step 6: Calculate the concentration of B at equilibrium:
Substitute the value of 'x' you obtained in step 5 into the equilibrium expression for B:
[B] = 2x
Therefore, once you solve for 'x' in step 5, you can calculate the concentration of B at equilibrium by multiplying 'x' by 2.