The reaction below has an equilibrium constant of
Kp=2.26×104 at 298 K.
CO(g)+2H2(g)⇌CH3OH(g)
Calculate Kp for the reaction below.
3CH3OH(g)⇌3CO(g)+6H2(g)
(1/226000)^3 = 8.66E-14
Answered above.
You made a typo; you have too many zeros in that 226000 don't you? The answers looks ok.
Well, calculating Kp is like trying to solve a puzzle, but I'm here to clown around and make it fun! Let's break it down step by step.
First, we need to understand that Kp is basically a measure of how much the reaction wants to go forward towards the products.
In the given reaction, we have 3 moles of CH3OH turning into 3 moles of CO and 6 moles of H2. So, we can say that the coefficient of CH3OH is 3, CO is 3, and H2 is 6.
Now, let's use the equation: Kp = (P(CO)^3 x P(H2)^6) / P(CH3OH)^3.
Since we're given Kp for the first reaction, which is 2.26×10^4, we can plug that in and rearrange the equation to solve for the unknown P(CH3OH)^3:
2.26×10^4 = (P(CO)^3 x P(H2)^6) / P(CH3OH)^3.
Solving for P(CH3OH)^3, we get:
P(CH3OH)^3 = (P(CO)^3 x P(H2)^6) / 2.26×10^4.
Now, to find the Kp for the second reaction, we need to multiply the original Kp by the concentration of each substance raised to the power of their coefficients. So, the final expression for the Kp of the second reaction is:
Kp = (P(CO)^3 x P(H2)^6) / [2.26×10^4]^3.
Remember, this is a simplified explanation, so have fun clowning around with the numbers and make sure to double-check your work!
To calculate the equilibrium constant for the reaction 3CH3OH(g) ⇌ 3CO(g) + 6H2(g) (let's call it Kp'), we can use the relationship between the equilibrium constants of the forward and reverse reactions.
If the equilibrium constant for the forward reaction (from left to right) is Kp, then the equilibrium constant for the reverse reaction (from right to left) is the reciprocal of Kp.
So, if Kp = 2.26×10^4 for the reaction CO(g) + 2H2(g) ⇌ CH3OH(g), then Kp' for the reverse reaction CH3OH(g) ⇌ CO(g) + 2H2(g) would be 1/Kp.
Thus, Kp' = 1/Kp = 1/(2.26×10^4) = 4.425×10^(-5).
But the given reaction is not the reverse of the reaction in question. We have the reverse reaction CH3OH(g) ⇌ CO(g) + 2H2(g), but we need to find the equilibrium constant Kp' for 3CH3OH(g) ⇌ 3CO(g) + 6H2(g).
Since the equilibrium constant is determined by the stoichiometric coefficients of the reaction, we can use the following relationship:
Kp' = (Kp)^n
Where n is the coefficient of the balanced equation for the reaction, which in this case is 3. So,
Kp' = (Kp)^3 = (4.425×10^(-5))^3
By raising Kp to the power of 3, we get:
Kp' ≈ 8.398×10^(-15)
So, the equilibrium constant Kp for the reaction 3CH3OH(g) ⇌ 3CO(g) + 6H2(g) is approximately 8.398×10^(-15).