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)
Kp for the rxn is 2.26E4
For the reverse it is 1/Kp
For 3x the rxn it is (1/Kp)^3
8.66x10^-14
Oh, calculating equilibrium constants? That's easy-peasy lemon squeezy. Let's do this!
This reaction is just the reverse of the previous one, so we can use the inverse of its equilibrium constant. You know, turn it upside down and inside out, like a clown doing acrobatics! 🤹♂️
So, to calculate Kp for this reaction, we simply take the reciprocal of the previous equilibrium constant, Kp = 1/Kp_previous = 1/(2.26×10^4).
Now, I don't have my calculator handy, but I bet you do! Just plug in the numbers and crunch away. And don't worry, I guarantee that the answer will be clown-approved! 🤡
To calculate the equilibrium constant for the given reaction, we can use the concept of the equilibrium constant expression and the relationship between the reaction coefficients.
The given equation is: 3CH3OH(g) ⇌ 3CO(g) + 6H2(g)
First, we need to identify the changes in the coefficients compared to the original reaction, which is the reverse of the given equation. In this case, the coefficients are multiplied by 3 for each species.
From the original reaction:
CO(g) + 2H2(g) ⇌ CH3OH(g)
From the reverse reaction:
CH3OH(g) ⇌ CO(g) + 2H2(g)
Now, to calculate the equilibrium constant (Kp) for the reverse reaction, we need to square the original Kp value because the stoichiometric coefficients of the reactant and product are multiplied by 2:
Kp(reverse reaction) = (Kp(original reaction))^2
Plugging in the given Kp value: Kp(original reaction) = 2.26 × 10^4
Kp(reverse reaction) = (2.26 × 10^4)^2
Kp(reverse reaction) = 5.1076 × 10^8
Thus, the equilibrium constant (Kp) for the reaction 3CH3OH(g) ⇌ 3CO(g) + 6H2(g) is 5.1076 × 10^8.