N2-3H2- 2NH31 mol of N2 and 4 mol of H2 are taken in a 15 L flask at 27-C- After the complete conversion of N2 into NH3- 5 L of H2O is added- The pressure set up in the flask is-

Nbsp; amp;N_2 nbsp; amp;+ nbsp; amp;3H_2 nbsp; amp;→ nbsp; amp;2NH_3 nbsp; nbsp;t=0 amp;1 mole nbsp; amp; nbsp; amp;4 mo ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Chemistry

Ideal Gas Equation

N2+3H2→ 2NH31...

Question

$N_{2} + 3 H_{2} \to 2 N H_{3}$
1 mol of $N_{2}$ and 4 mol of $H_{2}$ are taken in a $15 L$ flask at $27^{\circ} C$ . After the complete conversion of $N_{2}$ into $N H_{3}$ , $5 L$ of $H_{2} O$ is added. The pressure set up in the flask is:

\frac{3 \times 0.0821 \times 300}{15} atm

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

\frac{2 \times 0.0821 \times 300}{10} atm

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

\frac{1 \times 0.0821 \times 300}{15} atm

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

\frac{1 \times 0.0821 \times 300}{10} atm

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

Open in App

Solution

The correct option is D $\frac{1 \times 0.0821 \times 300}{10} atm$
$\begin{matrix} N_{2} & + & 3 H_{2} & \to & 2 N H_{3} t = 0 & 1 mole & 4 mole & 0 t = t_{f i n a l} & 0 & 1 mole & 2 mole \end{matrix}$

$N H_{3}$ will be absorbed by water and the volume will be $15 - 5 = 10 L$ .

$P = \frac{n R T}{V} = \frac{1 \times 0.0821 \times 300}{10} atm$

Suggest Corrections

Similar questions

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

1 mol of

N_{2}

and 4 mol of

H_{2}

are taken in a

15 L

flask at

27^{\circ} C

. After the complete conversion of

N_{2}

into

N H_{3}

5 L

H_{2} O

is added. The pressure set up in the flask is:

2

mol of

N_{2}

and

3

mol

H_{2}

gas are allowed to react in a

20 L

flask at

400 K

and after complete conversion of

H_{2}

into

N H_{3}

10 L

H_{2} O

was added and temperature reduced to

300 K

. Pressure of the gas after reaction.

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

Q. Consider the following reaction equilibrium

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

. Initially,

1 m o l

N_{2}

and

3 m o l

H_{2}

are taken in a

2

litre flask. At equilibrium state if the number of mol of

N_{2}

0.6

what is the total number mol of

N_{2}

0.6

what is the total number mol of all gases present in the flask.

Q. A flask of capacity 1 L contains

N H_{3}

at 1 atm and

25^{\circ} C

. A spark is passed through until all the

N H_{3}

is decomposed to

N_{2}

and

H_{2}

. Calculate the pressure (in atm) of gases left at

25^{\circ} C

after the reaction.

Q. Consider the following reaction equilibrium:

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

. Initially, 1 mole of

N_{2}

and 3 moles of

H_{2}

are taken in a 2 litre flask.

At equilibrium state, if the number of moles of

N_{2}

is 0.6, then the total number moles of all gases present in the flask are:

Join BYJU'S Learning Program

N2+3H2→2NH3 1 mol of N2 and 4 mol of H2 are taken in a 15 L flask at 27∘C. After the complete conversion of N2 into NH3, 5 L of H2O is added. The pressure set up in the flask is:

The correct option is D 1×0.0821×30010 atmN2+3H2→2NH3t=01 mole4 mole0t=tfinal01 mole2 mole NH3 will be absorbed by water and the volume will be 15−5=10 L. P=nRTV=1×0.0821×30010 atm

$N_{2} + 3 H_{2} \to 2 N H_{3}$
1 mol of $N_{2}$ and 4 mol of $H_{2}$ are taken in a $15 L$ flask at $27^{\circ} C$ . After the complete conversion of $N_{2}$ into $N H_{3}$ , $5 L$ of $H_{2} O$ is added. The pressure set up in the flask is: