consider the reaction.\n2hf(g)⇌h₂(g)+f₂(g)\nat equilibrium at 600 k, the concentrations are as…

consider the reaction.\n2hf(g)⇌h₂(g)+f₂(g)\nat equilibrium at 600 k, the concentrations are as follows.\nhf=5.82×10⁻² m\nh₂=8.4×10⁻³ m\nf₂=8.4×10⁻³ m\nwhat is the value of k_eq for the reaction expressed in scientific notation?\n2.1×10⁻²\n2.1×10²\n1.2×10³\n1.2×10⁻³

consider the reaction.\n2hf(g)⇌h₂(g)+f₂(g)\nat equilibrium at 600 k, the concentrations are as follows.\nhf=5.82×10⁻² m\nh₂=8.4×10⁻³ m\nf₂=8.4×10⁻³ m\nwhat is the value of k_eq for the reaction expressed in scientific notation?\n2.1×10⁻²\n2.1×10²\n1.2×10³\n1.2×10⁻³

Answer

Explanation:

Step1: Write equilibrium - constant expression

The equilibrium - constant expression ($K_{eq}$) for the reaction $2HF(g)\rightleftharpoons H_2(g)+F_2(g)$ is $K_{eq}=\frac{[H_2][F_2]}{[HF]^2}$.

Step2: Substitute the given concentrations

Substitute $[HF] = 5.82\times10^{-2}\ M$, $[H_2]=8.4\times10^{-3}\ M$, and $[F_2]=8.4\times10^{-3}\ M$ into the expression: [ \begin{align*} K_{eq}&=\frac{(8.4\times 10^{-3})\times(8.4\times 10^{-3})}{(5.82\times 10^{-2})^2}\ &=\frac{8.4\times8.4\times10^{-3 - 3}}{5.82\times5.82\times10^{-2\times2}}\ &=\frac{70.56\times10^{-6}}{33.8724\times10^{-4}}\ &=\frac{70.56}{33.8724}\times10^{-6 + 4}\ &\approx2.1\times10^{-2} \end{align*} ]

Answer:

$2.1\times 10^{-2}$