The reaction has an equilibrium constant of Kp=2.26x10^4 at 298 K.
CO(g) + 2H_2(g) <-> CH_3OH(g)
Calculate Kp for the reactions and predict whether reactants or products will be favored at equilibrium.
A. CH_3OH(g) <-> CO(g) + 2H_2(g)
B. 1/2CO(g) + H_2(g) <-> 1/2CH_3OH(g)
A. The reaction for A is just the reverse of the original; therefore, the new Kp(that is K'p) for the new rxn will be 1/Kp
B. B is just 1/2 the original; therefore, the new Kp (that's K"p) will be sqrt(Kp)
For any reaction of A ==> B, then
Kp = pB/pA. So when Kp is a large number that means pB is large and pA is small so B (products) are favored.
When Kp is a small number that mans pB is small and pA is large so A (the reactants) are favored.
To calculate the equilibrium constant Kp for a reaction, you first need to write the balanced chemical equation for the reaction. In this case, the reaction is:
CO(g) + 2H₂(g) ⇌ CH₃OH(g)
To calculate Kp, we need to determine the stoichiometric coefficients of the balanced equation. In this case, the coefficients are 1 for CO, 2 for H₂, and 1 for CH₃OH.
Now, let's calculate Kp for the given reactions:
A. CH₃OH(g) ⇌ CO(g) + 2H₂(g)
To calculate Kp for the reverse reaction, we take the reciprocal of the Kp value for the forward reaction. So, Kp for the reverse reaction (A) is:
Kp(reverse_A) = 1 / Kp(forward)
Kp(reverse_A) = 1 / (2.26 * 10^4) = 4.42 * 10^(-5)
From the calculated Kp value, we can determine whether reactants or products will be favored at equilibrium. When Kp is a large number, it indicates that products are favored at equilibrium. Conversely, when Kp is a small number, it indicates that reactants are favored at equilibrium.
In this case, the calculated Kp value for the reverse reaction (A) is small (4.42 * 10^(-5)), suggesting that the reactants (CH₃OH) will be favored at equilibrium.
B. 1/2CO(g) + H₂(g) ⇌ 1/2CH₃OH(g)
To calculate Kp for this reaction, we need to consider that the stoichiometric coefficients are halved for both CO and CH₃OH.
Kp(reverse_B) = 1 / Kp(forward)
Kp(reverse_B) = 1 / (2.26 * 10^4) = 4.42 * 10^(-5)
Similar to the previous case, the calculated Kp value for the reverse reaction (B) is small, indicating that the reactants (1/2CH₃OH) will be favored at equilibrium.
To summarize:
A. For the reaction CH₃OH(g) ⇌ CO(g) + 2H₂(g), the reactants (CH₃OH) will be favored at equilibrium.
B. For the reaction 1/2CO(g) + H₂(g) ⇌ 1/2CH₃OH(g), the reactants (1/2CH₃OH) will be favored at equilibrium.
To calculate the equilibrium constant, Kp, for the given reactions, we'll first use the relationship between Kp values for the forward and reverse reactions:
Kp_reverse = 1 / Kp_forward
Now let's solve for Kp for each reaction and determine whether reactants or products will be favored at equilibrium:
A. CH3OH(g) <--> CO(g) + 2H2(g)
The given equilibrium constant (Kp) is for the forward reaction, so Kp_forward = 2.26 × 10^4.
Using the relationship mentioned earlier:
Kp_reverse = 1 / Kp_forward
Kp_reverse = 1 / (2.26 × 10^4)
Kp_reverse = 4.42 × 10^-5
Since Kp_reverse is a small value, this indicates that the reverse reaction (reactants) will be favored at equilibrium. Therefore, in reaction A, reactants will be favored at equilibrium.
B. 1/2CO(g) + H2(g) <--> 1/2CH3OH(g)
The given equilibrium constant (Kp) is for the forward reaction, so Kp_forward = 2.26 × 10^4.
Using the relationship mentioned earlier:
Kp_reverse = 1 / Kp_forward
Kp_reverse = 1 / (2.26 × 10^4)
Kp_reverse = 4.42 × 10^-5
Again, since Kp_reverse is a small value, this indicates that the reverse reaction (reactants) will be favored at equilibrium. Therefore, in reaction B, reactants will be favored at equilibrium.
In summary:
A. Reactants (CO and H2) will be favored at equilibrium.
B. Reactants (1/2CO and H2) will also be favored at equilibrium.