2H2O(g) ⇌ 2H2(g) + O2(g) K = 2.4×10^(-3) 2.0 L

2H2O(g) ⇌ 2H2(g) + O2(g) K = 2.4×10^(-3) 2.0 L

[H2O(g)] 1.1×10^(-1) M [H2(g)] 1.9×10^(-2) M

 

 

For the reaction 2H2O(g) ⇌ 2H2(g) + O2(g)

K = 2.4×10^(-3) at a given temperature.

At equilibrium in a 2.0 L container it is found that

[H2O(g)] = 1.1×10^(-1) M and [H2(g)] = 1.9×10^(-2) M.

Calculate the moles of O2(g) present under these conditions.

 

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K = ([H2]^2 [O2]) / [H2O]^2

 

2.4×10^(-3) = ((0.038)^2 (x)) / (0.22)^2

 

x = (2.4×10^(-3)) × (0.22)^2 / (0.038)^2

= 0.080 M = [O2]

---> O2의 농도는 0.080 M.

 

 

 

문제에서 요구하는 것은 O2의 몰수이므로,

(0.080 mol/L) × 2.0 L = 0.16 mol O2