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.
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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