Question

10 At high temperatures, steam decomposes into its elements according to the following equation.
2H2O(g) 2H2(g) + O2(g)
In one experiment at 1 atm pressure, it was found that 20% of the steam had been converted into
hydrogen and oxygen.
What are the values of the equilibrium partial pressures, in atm, of the components of this
equilibrium?

Answer 1

\[1 atm= P _{T}= P _{(H _{2}O)}+P _{(H_{2})}+ P _{(O_{2})}\]
\[P _{H _{2}O}=X_{H _{2}O}\times P_{T}\]
\[X_{H_{2}O}=\frac{ n _{H_{2}O} }{n_{T} }\]
the partial pressure of each component is going to be equal to the molar fraction of the component times the total pressure. You have to find out the composition of your final mix.

according to the reaction and the data in the problem said that 20% of the steam is converted, assuming that you have in the beginning 100 moles of H2O(g) and no H2 neither O2. You do and ICE table to find the composition of the final mix

2H2O(g) -> 2H2(g) + O2(g)
I 100 0 0
C -20 + 20 +10
E 80 20 10

the molar fraction for each component will be the number of moles of the component divided by the total number of moles in the mix (110)

H2O= 80/110
H2=20/110
O2=10/110

then you have to multiply the molar fraction of each component by the total pressure to calculate the partial pressure of each component in the mix