Question

Ammonium carbamate solid \(\left(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\right)\) decomposes at \(313 \mathrm{~K}\) into ammonia and carbon dioxide gases. At equilibrium, analysis shows that there are \(0.0451 \mathrm{~atm}\) of \(\mathrm{CO}_{2}\), 0.0961 atm of ammonia, and \(0.159 \mathrm{~g}\) of ammonium carbamate. (a) Write a balanced equation for the decomposition of one mole of \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\). (b) Calculate \(K\) at \(313 \mathrm{~K}\).

Short answer

Answer: The equilibrium constant (K) for the decomposition of ammonium carbamate at 313 K is approximately 1.63.

Step-by-step solution

Write the balanced equation for decomposition of NH4CO2NH2.

The given decomposition reaction is: NH4CO2NH2 (s) -> NH3 (g) + CO2 (g)

Calculate the moles of NH4CO2NH2 at equilibrium.

We know the mass of ammonium carbamate at equilibrium is 0.159 g. To find the number of moles, divide the mass by its molar mass. Molar mass of NH4CO2NH2 = 1(14.01 + 1.01) + 1(12.01) + 2(16.00) + 1(14.01 + 1.01) = 60.06 g/mol n(NH4CO2NH2) = (0.159 g) / (60.06 g/mol) = 0.00265 mol

Use the ideal gas law to calculate moles of NH3 and CO2 at equilibrium.

Because both NH3 and CO2 are gases, we can use the ideal gas law (PV=nRT) to find the moles for NH3 and CO2 at equilibrium: n(NH3) = (0.0961 atm) * V / (0.0821 L*atm/mol*K) (313 K) n(CO2) = (0.0451 atm) * V / (0.0821 L*atm/mol*K) (313 K) Since NH3 and CO2 are produced in a 1:1 ratio, their moles are equal: n(NH3) = n(CO2) = x moles Divide the pressure of NH3 by the pressure of CO2, and we get: V = x * (0.0821 L*atm/mol*K) (313 K) / 0.0961 atm Plug V back into the equation for n(CO2) to find x: x = (0.0451 atm) * [(x * (0.0821 L*atm/mol*K) (313 K)) / 0.0961 atm] / (0.0821 L*atm/mol*K) (313 K) x ≈ 0.00265 mol

Calculate the equilibrium constant, K

Now we can use the balanced equation and the values obtained in previous steps to find K. Since the balanced equation has a 1:1 ratio between NH3, CO2, and NH4CO2NH2, we can simplify the expression for K: K = [NH3] [CO2] / [NH4CO2NH2] K = (0.0961 atm) (0.0451 atm) / 0.00265 Calculate K: K ≈ 1.63 So, at 313 K, the equilibrium constant K for the decomposition of ammonium carbamate is approximately 1.63.