- #1
Nicci
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- Homework Statement
- The following equilibrium reaction is known as the 'water gas shift' reaction:
$$CO(g) + H_2O(g) \Leftrightarrow H_2(g) + CO_2(g)$$
When initially 0.40mol of CO(g) and 1.00 mol of $$H_2O(g)$$ are present in a vessel at $$1200°C$$, the pressure is 2 atm. The amount of $$CO_2(g)$$ is 0.225 mol at equilibrium.
a) Calculate the total equilibrium amount, n, of the gas mixture. Hint: Use the ICE table
b) Calculate the extent of dissociation, α.
c) Calculate the number of moles of each gas, except the carbon dioxide gas in the mixture at equilibrium.
- Relevant Equations
- pV=nRT
This is how I started my ICE table:
I am not sure if my table is correct. When I work out the total equilibrium amount, n, using the 1.4α I get: $$n = 0.40 - 0.40α +1.00 - 1.00α +1.4α +0.225$$
and that gives $$n = 1.625 mol$$
Rounding to the correct significant figures: n = 1.6 mol
I then used the 1.6 mol to get the extent of dissociation, α, but my answer does not make any sense when I substitute it back into the 'n' equation.
Can anyone give me hint to help me to see where I am making a mistake?
Thank you very much.
$$CO(g)$$ | $$H_2O(g)$$ | $$H_2(g)$$ | $$CO_2(g)$$ | |
Initial (mol) | 0.40 | 1.00 | 0 | 0 |
Change (mol) | -0.40α | -1.00α | 1.4α | 1.4α |
Equilibrium (mol) | 0.40-0.40α | 1.00-1.00α | 1.4α | 1.4α (or should I use 0.225) |
and that gives $$n = 1.625 mol$$
Rounding to the correct significant figures: n = 1.6 mol
I then used the 1.6 mol to get the extent of dissociation, α, but my answer does not make any sense when I substitute it back into the 'n' equation.
Can anyone give me hint to help me to see where I am making a mistake?
Thank you very much.