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At 500 ºC, the decomposition of water into hydrogen and oxygen, 2H2O(g) 2H2(g) + O2(g) has...

At 500 ºC, the decomposition of water into hydrogen and oxygen, 2H2O(g) 2H2(g) + O2(g) has Kc = 6.0 × 10-28. What are the concentrations of H2 and O2 that are present at equilibrium in a 6.00 L reaction vessel at this temperature if the container originally held 0.051 mol H2O? Enter your answer in scientific notation.

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Answer #1

Given concentration of water = 0.051 M

Concentration of water in reaction vessel = 0.051/6 = 0.0085 M

Chemical equation for the decomposition of water:

           2H2O    →   2H2   +      O2

            0              0         0

     (0.0085-2x)       (2x)      (x)

x = concentration of oxygen at equlibrium

2x= concentration of hydrogen at equilibrium

Given Kc = 6.0*10-28

Kc = [(2x)2*(x)]/(0.0085-2x)

(6.0*10-28)*(0.0085-2x) = 2x3

On simplification of this equation you will get the value of 'x'.

Getting 'x' value is the concentration of oxygen [O2] and '2x' is the concentration of hydrogen [H2]

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