ThermodynamicsHard

Question

Ethyl chloride(C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :
       C2H4 (g) + HCl (g) → C2H5Cl (g)        ᐃH = - 72.3 kJ/mol.
What is the value of ᐃE (in kJ), if 98 g of ethylene and 109.5 g of HCl are allowed to react at 300 K.

Options

A.- 64.81
B.- 190.71
C.- 209.41
D.- 224.38

Solution

ᐃng = 1–2 = –1
ᐃE = ᐃH - ᐃng RT = ᐃH - RT = - 72.3 + 8.314 × 300 × 10-3
= - 69.806 kJ/mole
so for 3 mole we will get ᐃE = × 69.806 × 3 kJ/mole = 209.42 kJ/mole

Create a free account to view solution

View Solution Free
Topic: Thermodynamics·Practice all Thermodynamics questions

More Thermodynamics Questions