In the equilibrium system N2(g) + 3H2(g) → 2NH3(g)Kc = 2.0 mol-2dm6 at 620 K. What is the value of Kp at this temperature.
Question
In the equilibrium system
N2(g) + 3H2(g) → 2NH3(g)
Kc = 2.0 mol<sup>-2</sup>dm<sup>6</sup> at 620 K.
What is the value of Kp at this temperature?
Solution
To find the value of Kp at 620 K, we need to use the relationship between Kc and Kp. The equation relating Kp and Kc is:
Kp = Kc(RT)^(∆n)
Where: Kp is the equilibrium constant in terms of partial pressures, Kc is the equilibrium constant in terms of molar concentrations, R is the ideal gas constant (0.0821 dm^3/mol·K), T is the temperature in Kelvin, ∆n is the difference in the number of moles of gaseous products and reactants.
In the given equation N2(g) + 3H2(g) → 2NH3(g), the difference in the number of moles of gaseous products and reactants is (∆n = 2 - (1 + 3)) = -2.
Now, we can substitute the values into the equation:
Kp = Kc(RT)^(∆n) Kp = 2.0 mol^-2 dm^6 * (0.0821 dm^3/mol·K * 620 K)^(-2)
Calculating this expression will give us the value of Kp at 620 K.
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