Consider the following reaction.
2A(aq) (Two way arrows) B(aq)
Kp = 5.48*10^-5 at 500K
If a sample of A at 3.10atm is heated to 500K, what is the pressure of B at equilibrium?
I know I need to do the ICE table, but am having trouble solving for x once I plug everything in.
"A" should equal 3.10-2x at equilibrium, and "B" should equal x. How do I solve for x at this point?
I have the equation:
x / (3.10-2x)^2 = 5.48x10^-5
Any help would be greatly appreciated! Thank you in advance!
........2A(aq) <--> B(aq)
I.......3.10........0
C.......-2x.........x
E.....3.10-2x.......x
5.48E-5 = (x)/(3.10-2x)
Clear the denominator.
5.48E-5(3.10-2x) = x
5.48E-5*3.10 - 5.48E-5*2x = x
Complete the multiplication, combine terms, solve for x.
To solve for the value of x in the equation x / (3.10-2x)^2 = 5.48x10^-5, you can follow these steps:
1. Expand the denominator: (3.10-2x)^2 = 9.61 - 12.4x + 4x^2
2. Multiply both sides of the equation by the denominator to eliminate it from the equation: x = (5.48x10^-5) * (9.61 - 12.4x + 4x^2)
3. Distribute the multiplication: x = (5.48x10^-5) * 9.61 - (5.48x10^-5) * 12.4x + (5.48x10^-5) * 4x^2
x = 5.30x10^-6 - 6.783x10^-5x + 2.192x10^-4x^2
4. Rearrange the equation to bring all terms to one side: 2.192x10^-4x^2 - 6.783x10^-5x + 5.30x10^-6 = 0
This gives you a quadratic equation in the form ax^2 + bx + c = 0.
In this case, a = 2.192x10^-4, b = -6.783x10^-5, and c = 5.30x10^-6.
Now, to solve for x, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Substitute the values of a, b, and c into the quadratic formula and solve for x.