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.
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
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 FreeMore Thermodynamics Questions
Which of the following processes takes place with decrease of entropy ?...In a given process on an ideal gas, dw = 0 and dq < 0. Then for the gas,...A cylinder with thermally insulated walls contains an insulated portion which can slide freely. The partition divides th...A student is calculating the work done by 2 mole of an ideal gas in a reversible isothermal expansion shown in the figur...The value of log10K for a reaction A ⇋ B is ( Given : áƒfHo298K = - 54.07 kJ mol-1, áƒrSo298K = 10JK-1mol-1 and R = ...