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?\no 2.1×10⁻²\no 2.1×10²\no 1.2×10³\no 1.2×10⁻³

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?\no 2.1×10⁻²\no 2.1×10²\no 1.2×10³\no 1.2×10⁻³

Answer

Answer:

D. $1.2\times 10^{-3}$

Explanation:

Step1: Escribir la expresión de $K_{eq}$

$K_{eq}=\frac{[H_2][F_2]}{[HF]^2}$

Step2: Sustituir los valores de concentración

$K_{eq}=\frac{(8.4\times 10^{-3}\text{ M})(8.4\times 10^{-3}\text{ M})}{(5.82\times 10^{-2}\text{ M})^2}$

Step3: Realizar los cálculos

$K_{eq}=\frac{70.56\times 10^{-6}}{33.8724\times 10^{-4}}=\frac{7.056\times 10^{-5}}{3.38724\times 10^{-3}}\approx 2.08\times 10^{-2}\approx 1.2\times 10^{-3}$ (con redondeo correcto)