Question

The value of kc for H2+CO2=H2O+ CO is 1.88 AT 1000c if 1 mole of H2 and cO2 are placed in 1LITER FLASK. The final equilibrium concentration of CO at 1000c is?

Answer 1

I make `[CO(g)]=0.578*mol*L^-1`

Explanation:

We address the equilibrium....

`H_2(g) + CO_2(g) stackrel(1000""^@C)rightleftharpoonsH_2O(g) + CO(g)`

And `K_c=([H_2O(g)][CO(g)])/([H_2(g)][CO_2(g)])=1.88`

And now initially....`[H_2]=[CO_2]=1*mol*L^-1`

And we ASSUME that `x*mol*L^-1` of dihydrogen, (and of course carbon dioxide) react....and so...we put values into our equilibrium expression..

`1.88=(x^2)/((1-x)^2)`

And thus `x/(1-x)=sqrt(1.88)=1.371`

`x=1.371-1.371x`

`2.371x=1.371`...and thus `x=0.578`

And so at equilibrium....`[H_2]=[CO_2]=1-0.578=0.422*mol*L^-1`...and of course `[H_2O(g)]=[CO(g)]=0.578*mol*L^-1`

Do the given concentrations give the required equilibrium constant? Please do not assume that I am infallible.