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 a 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! :)

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