28g of N2 and 6g of H2 were kept at 400 degree centigrade,the equilibrium mixture contained 27.54g of NH3. Find the value of Kc in mole^-2L^2 .
To find the value of Kc, we need to first write the balanced chemical equation for the reaction involved.
The balanced chemical equation for the reaction is:
N2(g) + 3H2(g) ⇌ 2NH3(g)
According to the equation, one mole of N2 reacts with three moles of H2 to produce two moles of NH3.
Next, we need to calculate the moles of N2, H2, and NH3 in the given masses.
Molecular mass of N2 = 28 g/mol
Moles of N2 = Mass of N2 / Molecular mass of N2 = 28 g / 28 g/mol = 1 mol
Molecular mass of H2 = 2 g/mol
Moles of H2 = Mass of H2 / Molecular mass of H2 = 6 g / 2 g/mol = 3 mol
Molecular mass of NH3 = 17 g/mol
Moles of NH3 = Mass of NH3 / Molecular mass of NH3 = 27.54 g / 17 g/mol = 1.62 mol
Now, we can use the balanced equation and the stoichiometry to determine the moles of N2, H2, and NH3 at equilibrium.
According to the balanced equation, we can see that for every 1 mole of N2 that reacts, 2 moles of NH3 are formed.
Since 1 mole of N2 reacts to form 2 moles of NH3, the number of moles of NH3 at equilibrium is twice the number of moles of N2 reacted.
Therefore, the number of moles of NH3 at equilibrium = 2 * Moles of N2 at equilibrium = 2 * 1.62 mol = 3.24 mol
Now we can write the expression for Kc using the equilibrium concentrations:
Kc = [NH3]^2 / ([N2] * [H2]^3)
Substituting the values:
Kc = (3.24 mol)^2 / (1 mol * (3 mol)^3) = 10.04 mol^-2 L^2
So, the value of Kc is 10.04 mol^-2 L^2.