Question

The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is increased from 2.00 liters to 8.00 liters, how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer.

co (g) + 3H2(g) -CH4(g) + H2O (g)

Answer 1

Q = 62.9

Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.

Step-by-step explanation:

Let's consider the following reaction.

CO(g) + 3 H₂(g) ⇄ CH₄(g) + H₂O(g)

The equilibrium constant (Kc) is:

`Kc= 3.93 =([CH_(4)].[H_(2)O])/([CO].[H_(2)]^(3) )`

If the volume is multiplied by 4 (2.00 L → 8.00 L), the concentrations will be divided by 4. The reaction quotient (Q) is:

`Q=((0.25[CH_(4)]).(0.25[H_(2)O]))/((0.25[CO]).(0.25[H_(2)])^(3) )=16([CH_(4)].[H_(2)O])/([CO].[H_(2)]^(3) )=16Kc=16 * 3.93 = 62.9`

Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.

Answer 2

the equilibium changes since Q > Kc, by increasing the volume, therefore, the reaction will try to use some of the excess product and favor the reverse reaction to reach equilibrium.

Step-by-step explanation:

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

∴ Kc = [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ] = 3.93.....equilibrium, V = 2.00L

PV = nRT; assuming P,T a standart conditions ( 1 atm, 298 K )

⇒ n / V = P / RT

∴ V = 8.00L

∴ R = 0.082 atm.L/K.mol

⇒ mol CO(g) = 0.327 mol = mol H2O = mol CH4

⇒ mol H2(g) = 0.327 mol CO * ( 3mol H2 / mol CO) = 0.981 mol

⇒ [ CO ] = 0.041 = [ CH4 ] = [ H2O]

⇒ [ H2 ] = 0.123 M

∴ Q = [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ]

⇒ Q = ( 0.041² ) / (( 0.123³ ) * ( 0.041 ))

⇒ Q = 22.03

• Q > Kc

⇒ we have more product present than we would have in the equilibrium.