Question

rates of reaction and equilibrium
consider the reaction below.
$h_{2}(g)+co_{2}(g)\to h_{2}o(g)+co(g)$
at equilibrium at 600 k, the following are true.
$co_{2}=9.5\times 10^{-4}m$
$h_{2}=4.5\times 10^{-2}m$
$h_{2}o=4.6\times 10^{-3}m$
$co=4.6\times 10^{-3}m$
what is the value of the equilibrium constant for this reaction in correct scientific notation?

Explanation:

Step1: Escribir la expresión de la constante de equilibrio

La constante de equilibrio \( K_c \) para la reacción \( H_2(g)+CO_2(g)
ightleftharpoons H_2O(g) + CO(g) \) está dada por \( K_c=\frac{[H_2O][CO]}{[H_2][CO_2]} \)

Step2: Sustituir los valores de concentración

Sustituimos \( [H_2O]=4.6\times 10^{-3}\,M \), \( [CO]=4.6\times 10^{-3}\,M \), \( [H_2]=4.5\times 10^{-2}\,M \) y \( [CO_2]=9.5\times 10^{-4}\,M \) en la fórmula de \( K_c \):

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Como \( \frac{10^{-6}}{10^{-6}} = 1 \), entonces \( K_c=\frac{21.16}{42.75}\approx0.495\approx4.95\times 10^{-1}\)

Answer:

\(4.9\times 10^{-1}\)