Could you please answer this question 1? Gas constant R = 8.315 JK^-1 mol^-1 = 0

Question

Could you please answer this question 1? Gas constant R = 8.315 JK^-1 mol^-1 = 0.08315 L bar K^-1 mol^-1; T(K) = t(degreeC) + 273.15 I atm = 101.325 kPa; 1 Torr = 133.3224 Pa: 1 bar = 10^5 Pa For reversible adiabatic changes, pV^gamma = constant with gamma = Cp,m/Cy,m = (Cv,m+ R)/Cv,m. A sample containing 2.0 mol of CaC03 was heated at 1073K until it completely decomposed to C02 and CaO. Calculate the heat of reaction from the following enthalpies of formation (assume ddeltafH/dT = 0): deltafH(CaCO3) = - 1207.1 kJ/mol deltafH(CaO) = - 653.09 kJ/mol deltafH(CO2) = - 393.51 kJ/mol Calculate the work done when the reaction is performed in a vacuum, Calculate the work done when .he reaction is performed m a container open to the atmosphere at 1 bar.

Explanation / Answer

CaCO3---> CaO+ CO2

Heat of reaction = enthalpy of products - enthalpy of reactants= -653.09-393.51-(-1207.1)= 160.5 Kh/mol

b) when the reactioons is carried out under vacuum it is the case of free expansion and hence the work done is zero

c) work done when the container is open the atmosphere = -Pdv= -P*nRT/P ( the initial volume is less compared to final volume)= -nRT =2*8.314* 1073K=-16844 joule

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