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

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

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

answer: the value of enthalpy change will be -152.80 kj.

solution:

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