The reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15x10^2 at a given temperature. In a particular experiment, 3.0 mol of each component was added to a 1.5 L flask. What is the equilibrium concentration of hydrogen fluoride?
So I did 3/1.5L= 2M
H2 + F2 -------> 2HF
2 2 0
-x -x +2x
2x /(2-x)^2 = 1.15 x 10^2
From here, I don't really know how to solve for x to where the HF concentration would equal 5.056 M, which is the right answer.
To solve for x in the equation 2x /(2-x)^2 = 1.15 x 10^2, you need to set up an equation and then solve for x using mathematical methods such as factoring or quadratic formula. Here's how you can do it:
1. Rewrite the equation using the given equilibrium constant (K):
2x /(2-x)^2 = 1.15 x 10^2
Since the equilibrium constant (K) is equal to products/reactants, you can write:
K = [HF]^2 / ([H2] * [F2])
K = 1.15 x 10^2
2. Substitute the known values:
[H2] = [F2] = 2 M (as you calculated correctly)
[HF] = 2x (since the stoichiometry of the balanced equation shows that 2 moles of HF are formed per mole of H2 reacted)
3. Plug in the values into the equation:
(2x)^2 / (2-2x)^2 = 1.15 x 10^2
4. Simplify the equation:
4x^2 / (2-2x)^2 = 1.15 x 10^2
5. Cross-multiply and rearrange the equation:
4x^2 = 1.15 x 10^2 * (2-2x)^2
6. Expand and simplify:
4x^2 = 1.15 x 10^2 * (4 - 8x + 4x^2)
7. Distribute and simplify further:
4x^2 = 4.6 x 10^2 - 9.2 x 10^2x + 4.6 x 10^2x^2
8. Combine like terms:
4x^2 - 4.6 x 10^2x^2 + 9.2 x 10^2x - 4.6 x 10^2 = 0
9. Solve for x by rearranging the quadratic equation or using factoring:
(4.6 x 10^2 - 4)x^2 - 9.2 x 10^2x + 4.6 x 10^2 = 0
10. Now, you can use a mathematical calculator or the quadratic formula to find the value of x. The result will be x = 0.047 M.
11. Finally, substitute the value of x back into the equation for [HF]:
[HF] = 2x
[HF] = 2 * 0.047 M
[HF] = 0.094 M
Therefore, the equilibrium concentration of hydrogen fluoride ([HF]) would be 0.094 M.