Grade 12th passPhysical Chemistry

0.2 mole of N2 and 0.6 mole of H2 react to give NH3 and 40% of reactant mixture is decreased,according to the equation
N2(g)+3H2(g)--->2NH3(g)
at constant temprature and pressure.then the ratio of final volume to initial volume of gases is??
a) 4:5
b) 5:4
c) 7:10
d) 8:5

shreeyans

10 Years agoGrade 12th pass

4 Answers

Suraj Prasad

10 Years ago

The answer is coming to be 7:10,
N2+ 3H2 → 2 NH3
0.2 0.6

0.2-x 0.6- 3x 2x

Acc to ques. (0.2-x)+ (0.6 -3x) = 0.4* 0.8

from here x coming out to be 0.12

final volume will be (0.08+0.24+0.24)= 0.56 moles

Initial volume is (0.2+0.6)= 0.8

Ratio = 0.56/0.8 = 7/10 = 7:10

I hope it helps :)

shreeyans

10 Years ago

but the correct answer is 4:5 ..i cross check from two book having same question and from internet also the correct answer is 4:5

Tushnik

9 Years ago

N2 + 3H2 > 2NH3(0.2-x). (0.6-3x). 2xNow .(0.2-x)+(0.6-3x)=60/100 *(0.2+0.6) Or. X=0.08Now total final mol =(0.2-0.08)+(0.6-3*0.08)+(2*0.08)=0.64And initial total mol=0.2+0.6=0.8Now. The ratio of mol=volume=0.64/0.8=4/5

Tushnik

9 Years ago

N2 + 3H2 > 2NH3(0.2-x) (0.6-3x) 2xNow (0.2-x)+(0.6-3x)=60/100 *(0.2+0.6) Or X=0.08Now total final mol =(0.2-0.08)+(0.6-3*0.08)+(2*0.08)=0.64And initial total mol=0.2+0.6=0.8Now The ratio of mol=volume=0.64/0.8=4/5(You have to notice 40 percent reacted and 60 percent left )