Question
Asked Oct 24, 2019

If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2(g) <--> 2NO(g) + O2(g) reaches equilibrium is 0.674 atm.  Calculate the equilibrium partial pressure of NO2.

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Expert Answer

Step 1

Given

The NO2 has initial partial pressure = 0.500atm

The reaction that is taking place

Ο,
2ΝΟ
2ΝΟ
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Ο, 2ΝΟ 2ΝΟ

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Step 2

The conditions as the reaction proceeds

2NO2 2NO+O2
0 0
initial conditions
0.5atm
equilibrium conditions
0.5-2x
2x
X
where
x is the partial pressure that the reactant gas loses
as the reaction proceeds
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2NO2 2NO+O2 0 0 initial conditions 0.5atm equilibrium conditions 0.5-2x 2x X where x is the partial pressure that the reactant gas loses as the reaction proceeds

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Step 3

Since the final partial pressur...

0.674atm- ((0.5-2x)+2x+x)atm
0.674atm (0.5 x)atm
(0.674-0.5)atm x
0.174atm-x
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0.674atm- ((0.5-2x)+2x+x)atm 0.674atm (0.5 x)atm (0.674-0.5)atm x 0.174atm-x

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