Question

Consider a chemical reaction CO(g) + 2H, (g) = CH OH(g). When 2 moles of Co were reacting with 4 moles of H, at temperature 500K and pressure p = 100 bar, and chemical equilibrium was reached, n = 0.366 moles of CO was consumed. a) (10 pts) Based on the given data determine the value of the chemical equilibrium constant K(T) for T = 500K. b) (10 pts) Consider now the same reaction, when 3 moles of CO are reacting with 3 moles of H, at the temperature T = 500K and the pressure p = 100 bar. Find the mass m of the product, i.e., CH,OH, obtained in the reaction.

Answer 1

a) For the given reaction, the stoichiometric relation tells us that the numbero of moles of H₂ that react is double the number of moles of CO that reacts. So, if 0.366 moles of CO are consumed to reach equilibrium, the number of moles of H₂ consumed is double this value: 0.732 moles. Also, since the number of moles of product obtained is the same as the number of moles of CO consumed, we can easily determine that 0.366 moles of product are formed in equilibrium. To calculate the concentrations of each species, we need the volume of the vessel, which can be calculated using the ideal gas formula with the information of the beginning of the reaction:

$pV=nRT$

$V=\frac{nRT}{p}=\frac{6moles\cdot 0.083\frac{L\cdot bar}{mol\cdot K}\cdot 500K}{100bar}=2.49L$

So, the concentrations in equilibrium are:

For CO:

$c=\frac{2moles-0.366moles}{2.49L}=0.656M$

For H₂:
$c=\frac{4moles-0.732moles}{2.49L}=1.31M$

And for CH₃OH:

$c=\frac{0.366moles}{2.49L}=0.147M$

With these values we can calculate the equilibrium constant at 500K as:

$K=\frac{[CH_{3}OH]}{[CO][H_{2}]^{2}}=\frac{0.147}{0.656\cdot 1.31^{2}}=0.131$

b) For this part, we need to follow the changes in concentration using an ICE chart (notice that, since we are starting with 6 moles total and the same conditions, the volume is still 2.49 L)

	[CO]	[H2]	[CH3OH]
initial	1.20 M	1.20 M	0
change	-x	-2x	+x
equilibrium	1.20 M -x	1.20 M - 2x	x

We can now write the equilibrium constant with the unknown value of x:

$K=0.131=\frac{x}{(1.20-x)(1.20-2x)^{2}}$

This is a cubic equation, which can be solved to yield two imaginary solutions and one real solution, which is clearly the one we are looking for:

x = 0.126 M

This is the concentration of methanol obtained, which is equivalent to a mass of (we are using the molar mass of methanol: 32.04 g/mol):

$m=M\cdot V(L)\cdot m_{molar}=0.126\frac{mol}{L}\cdot 2.49L\cdot 32.04\frac{g}{mol}=10.1g$