Use the following information to calculate the amount of heatinvolved in the com

Question

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

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