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veitch
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Homework Statement
CH4 (g) + H2O (g) <- -> CO (g) + 3 H2 (g)
.2 M CH4 and .2 M H2O are added to an empty 4 L sealed vessel at 1400 K. When equilibrium is established, the concentration of H2 is .444 M. Calculate Kc... which I did and got 4.8 which is the given answer.
The real problem I have is part b) What mass of H2O (g) must be added to the equilibrium mixture in order to raise the concentration of H2 to .54 mol/L? The answer is supposed to be "20. g"
The Attempt at a Solution
using the equilibrium concentrations I calculated for the first part as initial concentrations
CH4 H2O CO 3H2
I .052 .052 .148 .444
C -.032 -.032 +.032 +.096
E .02 .02 .18 .54
So to change the .444 M H2 to .53 M... the reactants would need a third more added to them? and .032 M/L in 4.0 L would be 0.128 mol... if the molar mass of H2O is 18.0152 then that would be 2.3 g... which is nowhere near 20 g
Am I way off? I'm confused.
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