For the equation, N2+3H2 = 2NH3 where the Kp is 4.3×10^-4 at 375°C. Calculate the Kc for the equation
The relationship between Kp and Kc is:
Kp = Kc(RT)^Δn
where R is the gas constant (0.0821 L·atm/mol·K), T is the temperature in Kelvin, and Δn is the difference between the sum of the stoichiometric coefficients of the products and the sum of the stoichiometric coefficients of the reactants.
For the equation N2 + 3H2 = 2NH3, Δn = (2 - 1) = 1.
At 375°C (648.15 K), we have:
Kp = 4.3×10^-4 = Kc(RT)^Δn
Kc = Kp/RT = (4.3×10^-4)/(0.0821)(648.15)^1
Kc = 5.8×10^-7
Therefore, the Kc for the equation N2 + 3H2 = 2NH3 is 5.8×10^-7 at 375°C.
To calculate the Kc for the equation N2+3H2 = 2NH3, we can use the equation:
Kp = Kc(RT)^(Δn)
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 L atm/(mol K)),
T is the temperature in Kelvin,
Δn is the change in the number of moles of gas between products and reactants.
In this case, the equation is:
N2 + 3H2 = 2NH3
Δn = (2-1) = 1 (since there is one less mole of gas on the product side compared to the reactant side).
Given information:
Kp = 4.3×10^-4
T = 375°C = 375 + 273.15 = 648.15 K
Using the equation for Kp and Kc mentioned earlier, we can rearrange it to solve for Kc:
Kc = Kp / (RT)^Δn
Substituting the given values:
Kc = (4.3×10^-4) / ((0.0821 L atm/(mol K)) * (648.15 K))^1
Calculating the value within the parentheses:
Kc = (4.3×10^-4) / (53.296 L atm/(mol K))^1
Therefore, the Kc for the equation N2+3H2 = 2NH3 at 375°C is approximately equal to 8.07×10^-6.