Question

2. You want to know the standard heat of reaction for the following reaction: C2H4 (g) + 2C12 (g) → CJHCİ3(l) + H2 (g) + HCl (g) Calculate the standard heat of reaction using the following reactions: a) C3H4 (g) + 2C12 (g) CthCl4 (l) + H2 (g) C2H2C4 (1) ->C2HCl (I) +HCI (g) where AH385.76 kJ/mol where AHr -35 kJ/mol b) Calculate the standard heat of reaction using the heats of formation from Table B.1, givern that the standard heat of formation for trichloroethylene is -276.2 kJ/mol. c) Calculate the standard heat of reaction using the heats of combustion from Table B given that the standard heat of combustion for trichloroethylene is -947.7 kJ/mol, for hydrogen chloride is -95.31 kJ/mol, and for chlorine gas is-169.23 kJ/mol d) Are the three standard heats of reaction the same? If not, how much do they differ? e If 300 mol/h of C2HCl (1) is produced, how much heat is transferred in the process? f) If the molar flow rate triples and the first reaction is doubled, how much heat is trans ferred in the process?

Answer 1

1. C2H4 (g) +2CL2 (g) → C2HCl3 (l) + H2 (g) + HCl (g)

C2H4 (g) + 2Cl2 (g) → C2H2Cl4 (I)+ H2 (g)   AHr -385.76 kJ/mol (1)

C2H2Cl4 (1) → C2HCl3 (1)+ HCI (g)   AHr -35 kJ/mol (2)

Adding the above equations we get the required equation

hence standard heat of reaction  of required equation is 350.76 KJ/mol

2. standard heat of reaction is equal to the sum of standard enthalpies of formation of products minus the reactants

therefore here  $\Delta$H reaction = $\Delta$H products - $\Delta$H reactants

$\Delta$H reactant =$\Delta$H (c2h4)+ $\Delta$H(CL2) =52.3 kj/mol +00 kj/mol

$\Delta$H products = $\Delta$H(C2HCL3) +$\Delta$H]Hcl) = -262.3 kj/mol +00 kj/mol -95.3 kj/mol= -357.6 kj/mol

$\Delta$H reaction = $\Delta$H products - $\Delta$H reactants= -357.6 -52.3 =-305.3 kj/mol

(calculated assuming the values at standard state, as no table is provided with the question )

3.$\Delta$ H reaction =$\Delta$ H products - $\Delta$H reactant

$\Delta$ H products =(-947.7-95.3-286 )kj/mol= -1329 kj/mol

$\Delta$H reactant =(-1411.11-169.23) kj/mol=-1580.34 kj/mol

$\Delta$H reaction = $\Delta$H products - $\Delta$H reactants= 251.34 kj/mol.

4. no all the heat are not same, they differ by about 20%.