solve.....class 11..........
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C2H4+3O2=2CO2+2H2O
∆H=enthalpy of formatiom of reaction is -1323 kj/mol
Now apply Hess law,
∆H+enthapy of ethelyne+enthalpy of 3o2= enthalpy of 2co2+enthalpy of 2h2o=2(-393.7)+(2*241.8)=-1271kj/mol
This means -1358.6 kJ of energy released in the formation of CO2 and H2O
In the LHS of the reaction one mole of C2H4 is formed first and then reacted with 3 moles of oxygen to give products. But the enthalpy of formation of O2 is given 0, hence∆h of reaction= -1323kj/mol
again,
∆H=(entalphy of c+ enthalpy of o2gas)*2+2(enthalpy of h2 +enthalpy of o2gas)
-1323=2*(-3.93.7)+(2*241.8)-(enthalpy of c2h2+ 3*0)
enthalpy of c2h2=1323-1271=52kj/mol.
∆H=enthalpy of formatiom of reaction is -1323 kj/mol
Now apply Hess law,
∆H+enthapy of ethelyne+enthalpy of 3o2= enthalpy of 2co2+enthalpy of 2h2o=2(-393.7)+(2*241.8)=-1271kj/mol
This means -1358.6 kJ of energy released in the formation of CO2 and H2O
In the LHS of the reaction one mole of C2H4 is formed first and then reacted with 3 moles of oxygen to give products. But the enthalpy of formation of O2 is given 0, hence∆h of reaction= -1323kj/mol
again,
∆H=(entalphy of c+ enthalpy of o2gas)*2+2(enthalpy of h2 +enthalpy of o2gas)
-1323=2*(-3.93.7)+(2*241.8)-(enthalpy of c2h2+ 3*0)
enthalpy of c2h2=1323-1271=52kj/mol.
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