Complete question:
Consider the reaction.
At equilibrium at 600 K, the concentrations are as follows.
2HF -----> H₂ + F₂
[HF] = 5.82 x 10-2 M
[H2] = 8.4 x 10-3 M
[F2] = 8.4 x 10-3 M
What is the value of Keq for the reaction expressed in scientific notation?
2.1 x 10-2
2.1 x 102
1.2 x 103
1.2 x 10-3
Answer:
2.1 × 10^-2
Explanation:
Kequilibrum(Keq) = product/reactant
Equation for the reaction :
2HF -----> H₂ + F₂
Therefore,
Keq = [H2][F2] / [HF]^2
Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2
Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]
Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]
Keq = 2.0665 × 10^(-6 - (-4))
Keq = 2.0665 × 10^(-6 + 4)
Keq = 2.1 × 10^-2
