Answer:
Kc =<u> 3.74*10⁻³
</u>
Kp = 25.21
Explanation:
Step 1: Data given
Temperature = 1000 K
Volume = 5.00 L
Mass of CO = 8.62 grams
Mass of H2 = 2.60 grams
Mass of CH4 = 43.0 grams
Mass of H2O = 48.4 grams
Kc = [CO]*[H₂]³ / ([CH₄]∙*H₂O])
Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )
Step 2: The balanced equation
CH₄ + H₂O ⇄ CO + 3 H₂
Step 3: Calculate number of moles
The number of moles of each compund in the equilibrium mixture are:
Moles = mass / molar mass
n(CH₄) = 43.0g / 16g/mol = 2.688mol
n(H₂O) = 48.4g / 18g/mol = 2.689mol
n(CO) = 8.62g/28g/mol = 0.308mol
n(H₂) = 2.60g / 2g/mol = 1.3mol
Step 4: Calculate concentrations at equilibrium
So the equilibrium concentrations are:
Concentration = moles / volume
[CH₄] = 2.688mol/5L = 0.5376 M
[H₂O] = 2.689mol/5L = 0.5378M
[CO] = 0.308mol/5L = 0.0616M
[H₂) = 1.3mol/5L = 0.26M
Step 5: Calculate Kc
Kc = 0.0616 ∙ (0.26)³ / (0.5376∙0.5378) = <u>3.74*10⁻³
</u>
Step 5: Calculate partial pressure
Partial pressures in equilibrium can be found from ideal gas law:
p(X) = n(X)∙R∙T/V = [X]∙R∙T
=> p(CH₄) = [CH₄]∙R∙T = 0.5376mol/L * 0.082 06Latm/molK ∙ 1000K = 44.11 atm
p(H₂O) = [H₂O]∙R∙T = 0.5738mol/L * 0.082 06Latm/molK * 1000K = 44.13 atm
p(CO) = [CO]∙R∙T = 0.0616mol/L * 0.082 06Latm/molK * 1000K = 5.05atm
p(H₂) = [CO]∙R∙T = 0.26mol/L * 0.082 06Latm/molK * 1000K = 21.34atm
Step 5: Calculate Kp
Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )
Kp = 5.05*21.34³ / (44.11*44.13 ) = 25.21
