Answer:
2.16x10⁻²
Explanation:
First, let's find out the molar concentrations of the reactants. The molar mass of CH4 is 16 g/mol, and of CO2 is 44 g/mol. The number of moles is the mass divided by the molar mass:
nCH4 = 24.0/16 = 1.5 moles
nCO2 = 88.0/44 = 2 moles
The concentration is the number of moles divded by the volume, thus:
[CH4] = 1.5/1 = 1.50 M
[CO2] = 2/1 = 2.00 M
For the equilibrium reaction, let's do an equilibrium chart:
CH4(g) + CO2(g) ⇄ 2CO(g) + 2H2(g)
1.50 2.00 0 0 Initial
-x -x +2x +2x Reacts (stoichiometry is 1:1:2:2)
1.50-x 2.00-x 2x 2x Equilibrium
As sateted in problem, [CH4] = 2.70*[CO]
1.50 - x = 2.70*2x
1.50 - x = 5.4x
6.4x = 1.50
x = 0.2344
Thus, at equilibrium:
[CH4] = 1.50 - 0.2344 = 1.2656 M
[CO2] = 2.00 - 0.2344 = 1.7656 M
[CO] = 2*0.2344 = 0.4688 M
[H2] = 2*0.2344 = 0.4688 M
The equilibrium constant is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients.
K = ([CO]²*[H2]²)/([CH4]*[CO2])
K = (0.4688²*0.4688²)/(1.2656*1.7656)
K = 0.0483/2.2345
K = 2.16x10⁻²
