Use the following information to calculate the amount of heatinvolved in the complete reaction of 7.60 g of carbon to formPbCO3(s) in reaction 4. Be sure to give the proper sign(positive or negative) with your answer. (1) Pb(s) + 1/2 O2 -> PbO(s) Hdegree rxn = -219 kJ (2) C(s) +O2(g) ->CO2(s) Hdegree rxn = -394 kJ (3) PbCO3(s) -> PbO(s) +CO2(g) Hdegree rxn = 86 kJ (4) Pb(s) + C(s) + 3/2 O2(g) -> PbCO3(s) _____ kJ Use the following information to calculate the amount of heatinvolved in the complete reaction of 7.60 g of carbon to formPbCO3(s) in reaction 4. Be sure to give the proper sign(positive or negative) with your answer. (1) Pb(s) + 1/2 O2 -> PbO(s) Hdegree rxn = -219 kJ (2) C(s) +O2(g) ->CO2(s) Hdegree rxn = -394 kJ (3) PbCO3(s) -> PbO(s) +CO2(g) Hdegree rxn = 86 kJ (4) Pb(s) + C(s) + 3/2 O2(g) -> PbCO3(s) _____ kJ
Explanation / Answer
(1) Pb(s) + 1/2 O2 -> PbO(s) H 1 = -219 kJ (2) C(s) +O2(g) ->CO2(s) H 2 = -394 kJ (3) PbCO3(s) -> PbO(s) +CO2(g) H 3 = 86 kJ (4) Pb(s) + C(s) + 3/2 O2(g) -> PbCO3(s) : H = ? Eq(4) can be obtained by Eq(1) + Eq(2) + reverse ofEq(3) So, H = H1 + H 2 + ( - H 3 ) =-219 KJ + ( - 394 KJ ) + ( -86 KJ) = - 699 KJ Pb(s) + C(s) + 3/2 O2(g) ->PbCO3(s) : H = -699 KJ Molar mass of C = 12 g For 12 g of Cagbon The change in enthalphy is -699 KJ For 7.6 g of Cagbon The change in enthalphy is XKJ X = ( -699 * 7.6 ) / 12 = -442.7 KJ