Answer:
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.
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
⇒ we have more product present than we would have in the equilibrium.
Answer:
Q = 62.9
Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.
Explanation:
Let's consider the following reaction.
CO(g) + 3 H₂(g) ⇄ CH₄(g) + H₂O(g)
The equilibrium constant (Kc) is:
[tex]Kc= 3.93 =\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }[/tex]
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:
[tex]Q=\frac{(0.25[CH_{4}]).(0.25[H_{2}O])}{(0.25[CO]).(0.25[H_{2}])^{3} }=16\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }=16Kc=16 \times 3.93 = 62.9[/tex]
Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.
