Consider the following reaction: CO( g )+2H2( g )CH3OH( g ) A reaction mixture i

Question

Consider the following reaction:
CO(g)+2H2(g)CH3OH(g)
A reaction mixture in a 5.15 L flask at a certain temperature contains 26.5 g CO and 2.34 g H2. At equilibrium, the flask contains 8.65 g CH3OH.

Part A

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Consider the following reaction:
CO(g)+2H2(g)CH3OH(g)
A reaction mixture in a 5.15 L flask at a certain temperature contains 26.5 g CO and 2.34 g H2. At equilibrium, the flask contains 8.65 g CH3OH.

Part A

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Explanation / Answer

Solution

Given data

Volume = 5.15 L

Initially 26.5 g CO , 2.34 g H2

8.65 g CH3OH at equilibrium

Lets first calculate the moles of the each reactant and product

Moles = mass / molar mass

Moles of CO = 26.5 g / 28.0 g per mol = 0.946 mol CO

Moles of H2 = 2.34 g / 2.02 g per mol = 1.16 mol H2

Moles of CH3OH =8.65 g / 32 g per mol = 0.27 mol CH3OH

Now lets calculate the molarity of the each species

Molarity = moles / Liter

[CO] = 0.946 mol / 5.15 L = 0.184 M

[H2] = 1.16mol /5.15 L = 0.225 M

[CH3OH] = 0.27 mol / 5.15 L =0.0524 M

now lets calculate the equilibrium concentrations of the CO and H2

CO(g)+2H2(g)CH3OH(g)

We know the initial concentrations of the CO and H2

     CO(g)    +    2H2(g)              CH3OH(g)

I   0.184           0.225                        0

C -x                  -2x                            +x

E 0.184-x         0.225-2x                  0.0524

Here value of x is 0.0524 M

So using this value lets calculate the equilibrium concentrations of the CO and H2

[CO] = 0.184 – x = 0.184 M – 0.0524 M = 0.1316 M

[H2] = 0.225 -2x = 0.225 – (2*.0524) = 0.1202 M

Now lets write the kc equation for the reaction

Kc =[CH3OH]/[CO][H2]2

Lets put the values in the formula

Kc = [0.0524]/[0.1316][0.1202]2

Kc = 27.56

So the Kc for the reaction is 27.56

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