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For the reaction 2 H2O(g) equilibrium reaction arrow 2 H2(g) + O2(g), K = 2.4 ✕ 10−3 at a given temperature. At equilibrium in a 2.7 L container, it is found that [H2O(g)] = 2.0 ✕ 10−1 M and [H2(g)] = 2.5 ✕ 10−2 M. Calculate the moles of O2(g) present under these conditions.

Respuesta :

Answer:

[tex]0.414[/tex] moles of [tex]O_{2}[/tex] will be produced.

Explanation:

As per the given the expression for [tex]K_{c}[/tex] will be as follows,

[tex]K_{c} = \frac{[H_{2} ]^2[O_{2} ]}{[H_{2}O ]^2 }[/tex]

[tex]2.4\times10^{-2} = \frac{[2.5\times10^{-2} ]^2[O_{2} ]}{[2\times10^{-1} ]^2 }[/tex]

[tex][O_{2} ]=\frac{[2.4\times10^{-3}][2\times10^{-1} ]^2 }{[2.5\times10^{-2} ]^2}[/tex]

[tex][O_{2}]= 1.5\times10^{-1}M[/tex]

Also, [tex][O_{2} ]= \frac{moles}{LItre}[/tex]

[tex][1.5\times10^{-1} ]= \frac{moles}{2.7}[/tex]

[tex]moles =1.5\times10^{-1}\times2.7L[/tex]

[tex]0.414[/tex] moles [tex]O_{2}[/tex]

Hence,  [tex]0.414[/tex] moles [tex]O_{2}[/tex] will be produced in the given reaction.

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