The Haber process for ammonia synthesis is exothermic:
N2(g) + 3H2(g) <==> 2NH3(g) ∆H° = -92 kJ
If the equilibrium constant Kc for this process at 500.°C is 6.0 ξ 10-2, what is its value at 300.°C?
Use the van't Hoff equation.
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To find the equilibrium constant (Kc) at 300°C for the Haber process, we can use the equation:
ln(K2/K1) = (ΔH°/R) * (1/T1 - 1/T2)
Where:
K1 is the equilibrium constant at the initial temperature (500°C)
K2 is the equilibrium constant at the final temperature (300°C)
ΔH° is the standard enthalpy change (-92 kJ)
R is the gas constant (8.314 J/(mol·K))
T1 is the initial temperature in Kelvin (500°C + 273 = 773 K)
T2 is the final temperature in Kelvin (300°C + 273 = 573 K)
Let's plug in the given values and calculate K2:
ln(K2/6.0 ξ 10-2) = (-92 kJ / (8.314 J/(mol·K))) * (1/773 K - 1/573 K)
Note: We converted the units of ΔH° to J/mol and temperatures to Kelvin.
Now, solve for ln(K2/6.0 ξ 10-2), and then exponentiate both sides to find K2/6.0 ξ 10-2:
K2/6.0 ξ 10-2 = e^((-92 kJ / (8.314 J/(mol·K))) * (1/773 K - 1/573 K))
Finally, rearrange the equation to solve for K2:
K2 = (6.0 ξ 10-2) * e^((-92 kJ / (8.314 J/(mol·K))) * (1/773 K - 1/573 K))
Now, substitute the values into the equation and calculate K2 using a calculator or software.
Note: Make sure the units are consistent, such as using J instead of kJ for ΔH°.
Keep in mind that the final answer should be in the same units as the equilibrium constant K1, which is dimensionless.