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When 1.50 mol CO_2 and 1.50 mol H_2 are placed in a 0.750-L container at 395^0 C, ...

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roshan-rox | (Level 1) Valedictorian

Posted May 22, 2013 at 7:12 AM via web

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When 1.50 mol CO_2 and 1.50 mol H_2 are placed in a 0.750-L container at 395^0 C, 

`CO_2(g) + H_2(g)harrCO(g) + H_2O(g)` 

If K_c = 0.802, What are the concentrations of each substance in equilibrium mixture?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted May 22, 2013 at 8:04 AM (Answer #1)

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                       `CO_2+H_2 harr CO+H_2O`

Initial mol            1.5     1.5           -          -

Reacted mol          -x      -x           x         x

Equilibrium         1.5-x   1.5-x        x         x

 

`K_c = {[CO][H_2O]}/{[CO_2][H_2]}`

 

For the equilibrium solution;

`[CO_2] [H_2] = (1.5-x)/0.75` mol/l `= (3(1.5-x))/4` (mol)/l

`[CO] = [H_2O] = x/0.75 ` mol/l `= (3x)/4` (mol)/l

 

Only liquids and solids are omitted in concentration. But here we have to take `H_2O ` since it is a gas.

`K_c = {[CO][H_2O]}/{[CO_2][H_2]}`

`K_c = ((3x)/4)^2/((3(1.5-x))/4)^2`

`K_c = x^2/(1.5-x)^2`

`0.802 = x^2/(1.5-x)^2`

`sqrt0.802 = +-x/(1.5-x)`

`sqrt0.802 = +x/(1.5-x)`

`sqrt0.802 = -x/(1.5-x)`

Solving the above will give you x = 0.66759 or x = -6.076

Since x>0 then x = 0.6679

 

So the final concentration would be;

`[CO_2] [H_2] = (1.5-x)/0.75 = 1.109(mol)/l`

`[CO] = [H_2O] = x/0.75 = 0.89 (mol)/l`

 

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