Trichloroethylene, a widely used degreasing solvent for machine parts, is produced in a two-step reaction sequence. ethylene is first chlorinated to yield tetrachloroethane (rxn i), which is dehydrochlorinated to form trichloroethylene (rxn ii): rxn i: latex: c_2h_4\left(g\right)+2cl_2\left(g\r ight)\: \rightarrow\: c_2h_2cl_4\left(l\right)+h_2\left(g \right) c 2 h 4 ( g ) + 2 c l 2 ( g ) → c 2 h 2 c l 4 ( l ) + h 2 ( g ) latex: \bigtriangleup h^\circ_r=-385.76\: kj â–ł h r â = â’ 385.76 k j rxn ii: latex: c_2h_2cl_4\left(l\right)\: \rightarrow\: c_2hcl_3\left(l\right)+hcl\left(g\r ight) c 2 h 2 c l 4 ( l ) → c 2 h c l 3 ( l ) + h c l ( g ) the standard heat of formation of liquid trichloroethylene is -276.2 kj/mol. use hess's law to calculate the standard heat of the net reaction: latex: c_2h_4(g) + 2cl_2(g) \rightarrow c_2hcl_3(l) + h_2(g) + hcl(g) c 2 h 4 ( g ) + 2 c l 2 ( g ) → c 2 h c l 3 ( l ) + h 2 ( g ) + h c l ( g ) if 307 mol/h of c2hcl3(l) is produced in the net reaction and the reactants and products are all at 25oc and 1 atm, how much heat is evolved in the process (kw)? (assume latex: \dot{q} = \delta\dot{h} q ë™ = î” h ë™ )

a) C₂H₄ (g) + 2Cl₂ → C₂H₂Cl₄ (l) + H₂ (g), ΔH°rxn = -385.76 kJ

ΔH°rxn = ∑ΔH°f (products) - ∑ΔH°f (reactants)

ΔH°rxn = [ΔH°f (C₂H₂Cl₄) + ΔH°f (H₂)] - [ΔH°f (C₂H₄) + 2ΔH°f (Cl₂)]

-385.76 = [ΔH°f (C₂H₂Cl₄) + 0] - [ 52.3 + 2 (0) ]

ΔH°f (C₂H₂Cl₄) = -385.76 + 52.3 = -333.46 kJ /mol

For second reaction:

C₂H₂Cl₄ (l) → C₂HCl₃ (l) + HCl (g)

ΔH°rxn = ΔH°f (C₂HCl₃ ) + ΔH°f (HCl) - ΔH°f (C₂H₂Cl₄)

= -276.2 + (-92.3) -(-333.46)

ΔH°rxn = -35.04 kJ/mol

b) C₂H₄ (g) + 2Cl₂ (g) → C₂HCl₃ (l) + H₂ (g) + HCl (g)

standard heat of reaction = ΔH°₁+ ΔH°₂

= -385.76 - 35.04 = -420.8 kJ/mol

c) Q = ΔH = -420.8 kJ/mol × 307 mol/hr = -129185 kJ/hr of heat evolved.

Standard heat of the reaction is -420.8 kJ/mol

35.9 kW

Explanation:

Hess's law allows determination of total enthalpy change in a process based on the sum of enthalpy change of steps:

To produce trichloroethylene (C₂HCl₃) the reactions are:

C₂H₄(g) + 2Cl₂(g) → C₂H₂Cl₄(l) + H₂(g)

ΔH°rxn = [ΔH°f (C₂H₂Cl₄) + ΔH°f (H₂)] - [ΔH°f (C₂H₄) + 2ΔH°f (Cl₂)]  = -385.76 kJ/mol.

As  ΔH°f (H₂) and ΔH°f (Cl₂) are 0:

ΔH°f (C₂H₂Cl₄) - ΔH°f (C₂H₄) = -385.76 kJ/mol

As ΔH°f (C₂H₄) is 52.30 kJ/mol:

ΔH°f (C₂H₂Cl₄) - 52.30 kJ/mol = -385.76 kJ/mol

ΔH°f (C₂H₂Cl₄) = -333.46 kJ/mol

And C₂H₂Cl₄(l) → C₂HCl₃(l) + HCl(g);   ΔHf C₂HCl₃(l) = -276.2 kJ/mol

Where:

ΔH°rxn = [ΔH°f (C₂HCl₃) + ΔH°f (HCl)] - [ΔH°f (C₂H₂Cl₄)]

Knowing: ΔHf C₂HCl₃(l) = -276.2 kJ/mol; ΔH°f (HCl) = -92.31 kJ/mol; ΔH°f (C₂H₂Cl₄) = -333.46 kJ/mol:

ΔH°rxn = -276.2 kJ/mol - 92.31 kJ/mol + 333.46 kJ/mol

ΔH°rxn = -35.1 kJ/mol

The sum of both reactions gives:

C₂H₄(g) + 2Cl₂(g) → C₂HCl₃(l) + H₂(g) + HCl(g). Using Hess's law, the standard heat of this reaction is:

ΔH = -385.76 kJ/mol - 35.1 kJ/mol = -420.8 kJ/mol

As the production of 1 mole of C₂HCl₃(l) gives -420.8 kJ of energy, 307 mol/h gives:

307 mol/h × (-420.8 kJ/mol) =  -129186 kJ/h × (1h / 3600s) = 35.9 kJ/s = 35.9 kW

I hope it helps!

i believe the answer is a.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'