You visited us 2 times! Enjoying our articles? Unlock Full Access!

Atomic Number and Mass Number

For the react...

Question

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?

3.125

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

31.25

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

6.25

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

None of these

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

Open in App

Solution

The correct option is C $6.25$ %
$\frac{r_{m i x}}{r_{S O_{2}}} = \sqrt{\frac{M_{S O_{2}}}{M_{m i x}}} = \sqrt{\frac{64}{M_{m i x}}} = 2$
$M_{M i x} = 16$

Initial number of moles
$2$
$0$
$0$
Equilibrium number of moles
$2 - 2 x$
$x$
$3 x$
Mole fraction
$\frac{2 - 2 x}{2 + 2 x}$
$\frac{x}{2 + 2 x}$
$\frac{3 x}{2 + 2 x}$
$\frac{17 (2 - 2 x) + 28 x + 6 x}{2 + 2 x} = 16$
$x = \frac{1}{16}$
The percentage of ammonia converted is $\frac{2 \times 100}{2 \times 16} = 6.25$ %

Suggest Corrections

Similar questions

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

%

N H_{3}

S O_{2}

67.2

44.8

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

N_{2 (g)} + 3 H_{2 (g)} \to 2 N H_{3 (g)}; △_{r} H^{0}

- 91.8 k J m o l^{- 1}

N H_{3}

2 N H_{3 (g)} \to N_{2 (g)} + 3 H_{2 (g)}, △_{r} H^{0}

{N H}_{3}

N_{2}

H_{2}

{2 N H}_{3} \to N_{2} + {3 H}_{2}

x % o f {N H}_{3}

{S O}_{2}

x

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

K_{C} = L^{2} {m o l}^{- 2}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

K_{C} = \frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}

Join BYJU'S Learning Program