For the reaction HCONH2(g) NH3(g) + CO(g), Kc = 4.84 at 400 K. If ΔH° for this reaction is 29 kJ/mol, find Kc at 500 K.
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To find Kc at 500 K, we can use the equation:
ln(K₂/K₁) = ΔH°/R * (1/T₁ - 1/T₂)
Where:
K₁ = initial equilibrium constant at T₁ = 400 K (given as 4.84)
K₂ = equilibrium constant at T₂ = 500 K (to be found)
ΔH° = standard enthalpy change (given as 29 kJ/mol)
R = gas constant = 8.314 J/(mol*K)
T₁ = initial temperature = 400 K
T₂ = final temperature = 500 K
Substituting the given values into the equation:
ln(K₂/4.84) = (29,000 J/mol) / (8.314 J/(mol*K)) * (1/400 K - 1/500 K)
Simplifying the expression:
ln(K₂/4.84) = 3,491. Page 5Actions
We can now solve for K₂ by taking the exponential of both sides:
K₂/4.84 = e^(3,491)
K₂ = 4.84 * e^(3,491)
Using a calculator, we can find the value of K₂.
To find Kc at 500 K, we need to use the Van't Hoff equation, which relates the change in equilibrium constant (ΔlogK) to the change in temperature (ΔT) and the enthalpy change (ΔH°):
ΔlogK = ΔH° / R * (1 / T2 - 1 / T1)
Where:
ΔlogK is the change in log of the equilibrium constant
ΔH° is the enthalpy change of the reaction
R is the gas constant (8.314 J/(mol·K))
T1 and T2 are the initial and final temperatures, respectively.
First, we need to calculate the ΔlogK value using the given temperature and enthalpy change. Let's substitute the values into the equation:
ΔlogK = (29 kJ/mol) / (8.314 J/(mol·K)) * (1 / (500 K) - 1 / (400 K))
Simplifying the expression:
ΔlogK = (29,000 J/mol) / (8.314 J/(mol·K)) * (0.0025 K-1)
ΔlogK = 11.273
Now, we can find the new equilibrium constant Kc2 at 500 K using the equation:
logKc2 = logKc1 + ΔlogK
Where:
logKc1 is the logarithm of the equilibrium constant at 400 K.
Since Kc1 was given as 4.84, we can calculate:
logKc2 = log(4.84) + 11.273
Using logarithmic properties, we find:
logKc2 = 0.6840 + 11.273
logKc2 ≈ 11.957
Finally, we can find Kc2 by taking the antilog (10 raised to the power of logKc2):
Kc2 = 10^11.957
Kc2 ≈ 8.28 x 10^11
Therefore, at 500 K, the equilibrium constant Kc for the reaction HCONH2(g) ⇌ NH3(g) + CO(g) is approximately 8.28 x 10^11.