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Byju's Answer

Standard XII

Chemistry

Hydrogen Peroxide

2 mol of N2...

Question

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

3 R \times \frac{300}{20}

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3 R \times \frac{300}{10}

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R \times \frac{300}{20}

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R \times \frac{300}{10}

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Solution

The correct option is C $R \times \frac{300}{10}$
As given,
$N_{2}$ left $= 1$ mol
$N H_{3}$ formed $= 2$ mol but dissolved in $H_{2} O$ ; hence pressure is due to $N_{2}$ only. Hence, volume of the flask is $10 L$ since $10 L$ $H_{2} O$ is added.
$∴ P = \frac{n}{V} R T = \frac{300 R}{10}$

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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:

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:

11 m o l e s

N_{2}

and

12 m o l e s

H_{2}

mixture reacted in

20 L

vessel at

800 K

. After equilibrium was reached,

6 m o l e

H_{2}

was present.

3.58 L

of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to

300 K

. What is the final pressure of gaseous mixture/? Neglect vapour pressure of liquid solution. Assume (i) all

N H_{3}

dissolved in water (ii) no change in volume of liquid (iii) no reaction of

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and

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300 K

(Given

R = 0.821 L a t m m o l^{- 1} K^{- 1}

)

$2.8 g$ of $N_{2}$ gas at $300$ K and $20$ atm was allowed to expand isothermally against a constant external pressure of $1$ atm. Calculate $W$ for the gas.

Q. Determine the amount of heat (in kcal) given off at constant volume when

0.5 m o l

N_{2}

1.5

mol

H_{2}

reacted ording to equation at

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N_{2} (g) + 3 H_{2} (g) \to {2 N H}_{3} (g); Δ_{r} H_{300} = - 380 k c a l / m o l

(Given: R=2 cal/ mol-K)

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2 mol of N2 and 3 mol H2 gas are allowed to react in a 20L flask at 400K and after complete conversion of H2 into NH3, 10L H2O was added and temperature reduced to 300K. Pressure of the gas after reaction.N2+3H2→2NH3

The correct option is C R×30010As given,N2 left =1 molNH3 formed =2 mol but dissolved in H2O; hence pressure is due to N2 only. Hence, volume of the flask is 10L since 10L H2O is added.∴P=nVRT=300R10

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

The correct option is C $R \times \frac{300}{10}$
As given,
$N_{2}$ left $= 1$ mol
$N H_{3}$ formed $= 2$ mol but dissolved in $H_{2} O$ ; hence pressure is due to $N_{2}$ only. Hence, volume of the flask is $10 L$ since $10 L$ $H_{2} O$ is added.
$∴ P = \frac{n}{V} R T = \frac{300 R}{10}$