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If 8.51×10-2 moles of SO3(g), 0.210 moles of SO2, and 0.354 moles of O2 are at equilibrium in a 15.4 L container at 1.25×103 K, the value of the equilibrium constant, K, is ____
Question
Answer: The value of equilibrium constant, K, is [tex]7.74 \times 10^{-4}[/tex].
Explanation:
The reaction equation is as follows.
[tex]2SO_{3}(g) \rightleftharpoons 2SO_{2}(g) + O_{2}(g)[/tex]
Now, the expression for equilibrium constant of this reaction is as follows.
[tex]K = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}[/tex]
Substitute the given values into above expression as follows.
[tex]K = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}\\= \frac{(\frac{0.210}{15.4})^{2} \times (\frac{0.354}{15.4})}{(\frac{8.51 \times 10^{-2}}{15.4})}\\= 7.74 \times 10^{-4}[/tex]
Thus, we can conclude that the value of equilibrium constant, K, is [tex]7.74 \times 10^{-4}[/tex].