A mixture of NH3g and N2H4g is placed in a sealed container at 27 -C- 0-5 atm- After complete decomposition of NH3 and N2H4 at 927 -C the total pressure is reached to 4-5 atm-The decomposition reactions are as follow-2NH3g - N2g - 3H2gN2H4g - N2g - 2H2gThe percentage of NH3 by mole is-

Let, NH_3 = x mol nbsp; N_2H_4 = y mol 2NH_3(g) → N_2 (g)+ 3H_2(g) x 0 0 0 x2 3x2 N_2H_4 (g)→ N_2 (g)+ 2H_2(g) ...

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Standard XIII

Chemistry

Characteristics of Equilibrium Constant

A mixture of ...

Question

A mixture of $N H_{3} (g)$ and $N_{2} H_{4} (g)$ is placed in a sealed container at $27^{\circ} C$ , $0.5 atm$ . After complete decomposition of $N H_{3}$ and $N_{2} H_{4}$ at $927^{\circ} C$ the total pressure is reached to $4.5 atm$ .
The decomposition reactions are as follow:
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g) N_{2} H_{4} (g) \to N_{2} (g) + 2 H_{2} (g)$
The percentage of $N H_{3}$ (by mole) is:

75 %

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25 %

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40 %

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60 %

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Solution

The correct option is A $75 %$
Let,
$N H_{3} = x mol$
$N_{2} H_{4} = y mol$
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g) x 00 0 \frac{x}{2} \frac{3 x}{2}$
$N_{2} H_{4} (g) \to N_{2} (g) + 2 H_{2} (g) y 00 0 y 2 y$
$Initial total mole = x + y Final total mole = 2 x + 3 y$
$\frac{2 x + 3 y}{x + y} = \frac{4.5 \times 300}{(0.5) \times 1200} \Rightarrow \frac{2 x + 3 y}{x + y} = \frac{9}{4} \Rightarrow x = 3 y ∴ Percentage of N H_{3} = 75 %$

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Similar questions

Q. A mixture of

N H_{3} (g)

and

N_{2} H_{4} (g)

is placed in a sealed container at

27^{\circ} C

0.5 atm

. After complete decomposition of

N H_{3}

and

N_{2} H_{4}

927^{\circ} C

the total pressure is reached to

4.5 atm

.
The decomposition reactions are as follow:

2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g) N_{2} H_{4} (g) \to N_{2} (g) + 2 H_{2} (g)

The percentage of

N H_{3}

(by mole) is:

Q. A mixture of

N H_{3} (g)

and

N_{2} H_{4} (g)

is placed in a sealed container at

27^{\circ} C

0.5 atm

. After complete decomposition of

N H_{3}

and

N_{2} H_{4}

927^{\circ} C

the total pressure is reached to

4.5 atm

.
The decomposition reactions are as follow:

2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g) N_{2} H_{4} (g) \to N_{2} (g) + 2 H_{2} (g)

The percentage of

N H_{3}

(by mole) is:

Q. A mixture of

N H_{3} (g)

and

N_{2} H_{4} (g)

is placed in a sealed container at

300 K

. The total pressure is

0.5 a t m

. The container is heated to

1200 K

at which time both substances decompose completely according to the equations

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

and

N_{2} H_{4} (g) \to N_{2} (g) + 2 H_{2} (g)

. After decomposition is complete, the total pressure at

1200 K

is found to be

4.5 a t m

. Find the mole

%

N_{2} H_{4}

in the original mixture:

Q. A mixture of

N H_{3} (g)

and

N_{2} H_{4} (g)

is placed in a sealed container at

300 K

. The total pressure is

0.5 a t m

. The container is heated to

1200 K

at which both gases decompose completely according to the equation:

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

and

N_{2} H_{4} (g) \to N_{2} (g) + 2 H_{2} (g)

After complete decomposition, the total pressure at

1200 K

is found to be

4.5 a t m

. Find the moles % of

N_{2} H_{4}

in the original mixture :

Q. A mixture of

{N H}_{3} (g)

and

N_{2} H_{4} (g)

is placed in a sealed container at 300 K .The total pressure of the gaseous mixture is 0.5atm . The container is heated to 1200 K at which both

{N H}_{3}

and

N_{2} H_{4}

decomposed completely according to the following equation.

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

N_{2} H_{4} (g) ⟶ N_{2} (g) + 2 H_{2} (g)

After decomposition is complete, the total pressure at 1200 K is found to be 4.5 atm, The mole % of

N_{2} H_{4}

in the initial mixture at 300 K was:

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The correct option is A 75%Let, NH3=x mol N2H4=y mol 2NH3(g)→N2(g)+3H2(g) x 0 0 0 x2 3x2 N2H4(g)→N2(g)+2H2(g) y 0 0 0 y 2y Initial total mole=x+yFinal total mole=2x+3y 2x+3yx+y=4.5×300(0.5)×1200⇒2x+3yx+y=94⇒x=3y∴Percentage of NH3=75%