Combustion of urea CO(NH2)2)is described by the equation: CO(NH2)2 (s) + 3/2 O2
ID: 874254 • Letter: C
Question
Combustion of urea CO(NH2)2)is described by the equation: CO(NH2)2 (s) + 3/2 O2 (g) CO2 (g) + N2 (g) + 2H2O(l) The standard enthalpy of combustion of urea is -632 kJ per mole of urea at 298 K. a) How much heat is either released or adsorbed per mole of oxygen consumed b) The molar entropy of the combustion is 104.60 J K^-1 per mole of urea. can you identify two distinct reasons why the combustion entropy is positive? What is the Gibbs free energy change of combustion per mole of urea at 298 K? Is it a spontaneous process? c) Calculate the standard Gibbs energy of formation per mole of urea at 298K. The formation Gibbs free energy of CO2 = -394 kJ/mole and that of liquid water is -237 kl/mole. d) Does increasing the temperature stimulate combustion? Same question if the pressure is increased.Explanation / Answer
CO(NH2)2(s) + 3/2 O2(g) --------> CO2(g) + N2(g) + 2H2O(l)
(a)
deltaHoC = deltaHfo (products) - deltaHfo (reactants)
- 632 kJ/mol = deltaHfo(CO2) + deltaHfo(N2) + 2 * deltaHfo(H2O) - 3/2 deltaHfo (O2) - deltaHfo (CO(NH2)2)
- 632 kJ/mol = - 393.5 kJ/mol - 0 - 2 * (285.8) kJ/mol - 0 - deltaHfo (CO(NH2)2) [ enthalpy of formation of elemental states = 0 ]
- 632 kJ/mol = - 393.5 - 571.6 kJ/mol - deltaHfo (CO(NH2)2)
deltaHfo (CO(NH2)2) = ( - 965.1 + 632) kJ/mol
deltaHfo (CO(NH2)2) = - 333.1 kJ/mol ( It is deltaHfo per mole of Urea)
So deltaHfo (CO(NH2)2) per mole of Oxygen = - 2 / 3 * 333.1 kJ/mol = - 222.07 kJ/mol
So heat is released because sign is negative and the value is 222.07 kJ/mol
(b)
deltaSo per mole of Urea = 104.6 J/K
Combustion entropy is positive because firstly the number of gaseous molecules increase on combustion
At the left hand side 1.5 mol of O2 but at the right hand side 1 mol of CO2 and 1 mol of N2 so two moles of gas. So increase in gas molecules increases the disorder and hence entropy is positive.
Also the availability of urea for combustion decreases as it has changed into products so lesser availability of products means higher disoreder and positive entropy.
deltaGo per mole of Urea = deltaHo per mole of Urea - T * deltaSo per mol of Urea
deltaHo per mole of Urea = - 333.1 kJ
deltaSo per mole of Urea = 104.6 J/K
So,
deltaGo = - 333.1 x 10^3 J - 298 K * 104.6 J/K
deltaGo = - 364.27 kJ
Since sign is negative hence the process is spontaneous
(c)
deltaGof = deltaGo (products) - deltaGo (reactants)
deltaGof = deltaGo (CO2) + 2 * deltaGo (H2O) - 0 ( because deltaGfo = 0 for elements)
deltaGof = - 394 kJ/mol + 2 * ( - 237 kJ/mol)
deltaGof = - 868 kJ/mol
(d)
Because heat is released in combustion process hence increasing the temperature will lead the reaction in reverse direction and combustion will decrease hence increasing temperature won't stimulate combustion.
On increasing the pressure the reaction will move to a direction where there is less number of gaseous products. Hence on increasing pressure the reaction moves in reverse direction and rate of combustion is decreased.