Consider the reaction below. H2(g) CO2(g) Right arrow. H2O(g) CO(g) At equilibrium at 600 K, the following are true. [CO2] = 9. 5 x 10-4 M [H2] = 4. 5 x 10-2 M [H2O] = 4. 6 x 10-3 M [CO] = 4. 6 x 10-3 M What is the value of the equilibrium constant for this reaction in correct scientific notation? 4. 9 x 10-3 4. 9 x 10-2 4. 9 x 10-1 4. 9 x 103.

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31-01-2022
Chemistry

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Consider the reaction below. H2(g) CO2(g) Right arrow. H2O(g) CO(g) At equilibrium at 600 K, the following are true. [CO2] = 9. 5 x 10-4 M [H2] = 4. 5 x 10-2 M [H2O] = 4. 6 x 10-3 M [CO] = 4. 6 x 10-3 M What is the value of the equilibrium constant for this reaction in correct scientific notation? 4. 9 x 10-3 4. 9 x 10-2 4. 9 x 10-1 4. 9 x 103.

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shrutiagrawal1798 shrutiagrawal1798 · Answer 1 · 2022-02-09T14:24:26+01:00

The value of the equilibrium constant for the reaction is [tex]4.9\;\times\;10^{-1}[/tex]. Thus, option C is correct.

The equilibrium constant for the reaction is given as the ratio of the concentration of product to reactant raised to the stoichiometric coefficients.

Computation for the equilibrium constant

The balanced equation for the reaction is:

[tex]\rm H_2\;+\;CO_2\;\to\;H_2O\;+\;CO[/tex]

The equilibrium constant (K) for the given reaction is:

[tex]K=\rm \dfrac{[H_2O]\;[CO]}{[H_2]\;[CO_2]}[/tex]

Substituting the values for the concentration of the reactant and product:

[tex]K=\dfrac{[4.6\;\times\;10^{-3}]\;[4.6\;\times\;10^{-3}]}{[4.5\;\times\;10^{-2}]\;[9.5\;\times\;10^{-4}]} \\\\K=0.49\\\\K=4.9\;\times\;10^{-1}[/tex]

The value of the equilibrium constant for the reaction is [tex]4.9\;\times\;10^{-1}[/tex]. Thus, option C is correct.

Learn more about the equilibrium constant, here:

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