CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

For the reaction $$2NH_3(g) \rightarrow  N_2(g) + 3H_2(g)$$, what is the $$\%$$ of $$NH_3$$ converted if the mixture diffuses twice as fast as that of $$SO_2$$ under similar conditions :


A
3.125%
loader
B
31.25%
loader
C
6.25%
loader
D
62.5%
loader

Solution

The correct option is D $$6.25\%$$
  $$\displaystyle NH_3  $$$$\displaystyle  N_2$$ $$\displaystyle  H_2$$
Initial moles
 10
0
 Equilibrium moles
 $$\displaystyle  1-2x$$$$\displaystyle  x$$
$$\displaystyle  3x$$
 Equilibrium mole fraction
 $$\displaystyle \frac {1-2x}{1+2x} $$ $$\displaystyle \frac {x}{1+2x} $$ $$\displaystyle \frac {3x}{1+2x} $$
The rate of diffusion is inversely proportional to the square root of molar mass.
The mixture diffuses twice as fast as that of $$SO_2$$ under similar conditions.
$$\displaystyle \frac {r}{r'} = \sqrt {\frac {M'}{M}} $$
$$\displaystyle \frac {2}{1} = \sqrt {\frac {64}{M}} $$
$$\displaystyle M = 16 $$  $$g/mol
$$
$$\displaystyle 16 =  \frac {17(1-2x) + 28(x) + 2(3x)}{1+2x} $$
$$\displaystyle  16+32x = 17-34x+28x+6x $$
$$\displaystyle  32x = 1$$
$$\displaystyle x = 0.03125 $$
The percentage of ammonia converted
$$\displaystyle \frac {2x}{1} \times 100 = \frac {2 \times 0.03125}{1} \times 100 = 6.25 $$%

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image