CameraIcon

SearchIcon

MyQuestionIcon

8

You visited us 8 times! Enjoying our articles? Unlock Full Access!

Compressibility Factor

NH3 converts ...

Question

${N H}_{3}$ converts into $N_{2}$ and $H_{2}$ according to the following equation:
${2 N H}_{3} \to N_{2} + {3 H}_{2}$
$x % o f {N H}_{3}$ is converted into product. Rate of diffusion of gaseous mixture is double than that of ${S O}_{2}$ under identical conditions. Calculate $x$ .

A

98.23 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

31.25 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

25.27 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

6.25 %

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $6.25 %$
According to the Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its Molar mass.

$%$ of $N H_{3}$ converted into product = $x$ .
$2 N H_{3} \to N H_{2} + 3 H_{2}$
$t = 0$ $2$ $0$ $0$
$t = e q u i l i b r i u m$ $2 - 2 x$ $x$ $3 x$

Total number of moles = $2 - 2 x + x + 3 x$

$M_{m i x} = \frac{(n_{N H_{3}}) \times M_{N H_{3}} + (n_{H_{2}}) \times M_{H_{2}} + (n_{H_{2}}) \times M_{H_{2}}}{T o t a l n u m b e r o f m o l e s}$

$M_{m i x} = \frac{34}{2 + 2 x}$

Now, $\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

$r_{S O_{2}} = \frac{1}{2} r_{m i x}$

$\frac{r_{S O_{2}}}{r_{m i x}} = \sqrt{\frac{M_{m i x}}{M_{S O_{2}}}}$
$M_{M i x} = 16$

Hence, $16 = \frac{17}{1 + x}$

$x = \frac{1}{16}$
$%$ of $x = 6.25 %$
Hence, the correct option is D.

Suggest Corrections

thumbs-up

0

Similar questions

Q. For the reaction

2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)

%

N H_{3}

converted if the mixture diffuses twice as fast as that of

S O_{2}

under similar conditions :

Q. For the reaction,

2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g),

What is the percentage of

N H_{3}

converted, if the mixture diffuses twice as fast as that of

S O_{2}

under similar conditions?

34

gm of a mixture containing

N_{2}

H_{2}

1 : 3

by mole is partially converted into

{N H}_{3}

. Calculate the vapour density of the mixture (containing remaining

N_{2}, H_{2}

{N H}_{3}

formed) after reaction if it has been found that the

{N H}_{3}

formed required

0.5

H_{3} {P O}_{4}

for complete neutralization.

Q. 34 gm of a mixture containi ng

N_{2}

H_{2}

in I : 3 by mole is partially converted into

N H_{3}

. Calculate the vapour density of the mixture (containing remaining

N_{2}

H_{2}

N H_{3}

after reaction if it has been found that the

N H_{3}

form required 0.5 moles of

H_{3} P O_{4}

for complete neutralization.

3 N H_{3} + H_{3} P O_{4} \to (N H_{4})_{3} P O_{4}

Q. For the reaction,

N_{2} + 3 H_{2} \to 2 N H_{3}

under certain conditions of P and T, the rate of formation of

N H_{3}

k g h r^{- 1}

. The rate of conversion of

H_{2}

under these conditions is

x \times 10^{- 4}

h r^{- 1}

100 x

is____________.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Equations Redefined

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Compressibility Factor

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon