Ammonium carbamate solid (NH4CO2NH2) decomposes at 313 K into ammonia and carbon

Question

Ammonium carbamate solid (NH4CO2NH2) decomposes at 313 K into ammonia and carbon dioxide gases. At equilibrium, analysis shows that there are 0.0451 atm of CO2, 0.0961 atm of ammonia, and 0.159g of ammonium carbamate....a) Write the balanced equation for the decomposition of one mole of NH4CO2NH2.....b) Calculate K at 313 K.

Explanation / Answer

A) Balanced Chemical Equation: NH4CO2NH2 (s) --> 2NH3 (g) + CO2 (g) B) First, solve for the moles of NH4CO2NH2 from its molar mass: 0.159 g x 1 mol/78.07 g = 0.00204 moles Now, solve for the moles of NH3 and CO2 from the moles of NH4CO2NH2 using the balanced chemical equation: 0.00204 moles of NH4CO2NH2 x 2 moles of NH3/1 mole of NH4CO2NH2 = 0.00408 moles of NH3 0.00204 moles of NH4CO2NH2 x 1 mole of CO2/1 mole of NH4CO2NH2 = 0.00204 moles of CO2 Now, solve for the volume of NH3 and CO2 using the ideal gas equation, PV = nRT: P of NH3 = 0.0961 atm n = 0.00408 moles (calculated above) R = 0.0821 atm*L/mol*K (gas constant) T = 313 K V = ? PV = nRT V = nRT / P V = [(0.00408 moles)(0.0821 atm*L/mol*K)(313 K)] / 0.0961 atm V = 1.09 L PV = nRT V = nRT / P V = [(0.00204 moles)(0.0821 atm*L/mol*K)(313 K)] / 0.0451 atm V = 1.16 L K = [NH3]^2 * [CO2] NOTE: solids are not included in the Keq constant K = (0.00408 moles/1.09 L)^2 * (0.00204 moles/1.16 L) K = 2.46 x 10^-8 Hope this helps! :)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.