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Standard X

Chemistry

Le Chatelier's Principle

How can be th...

Question

How can be the concentration of hydrazine, $N_{2} H_{4}$ , could be increased in an equilibrium described by the equation.

$N_{2} (g) + 2 H_{2} (g) ⇌ N_{2} H_{4} (g)$ $Δ H = 95 k J$

Add

N_{2}

and

H_{2}

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Increase the pressure

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Heat the reaction

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All of the above

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Solution

The correct option is D All of the above
$N_{2} (g) + 2 H_{2} (g) ⇌ N_{2} H_{4} (g)$ $Δ H = 95 k J$
1 1 1
So to increase product, we have to add reactant.
Also as no of molecules of product are less so by increasing pressure, product increases.
Also as this reaction endothermic so this is favored by heat.
Hence, according to lechatelier's principle All of the given options are correct

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Similar questions

Q. in which of the following ways the concentration of hydrazine,

N_{2} H_{4}

, could be increased in an equilibrium described by the equation

N_{2} (g) + 2 H_{2} (g) ⇌ N_{2} H_{4} (g) Δ H = 95 k J

Q. in which of the following ways the concentration of hydrazine,

N_{2} H_{4}

, could be increased in an equilibrium described by the equation :

N_{2} (g) + 2 H_{2} (g) ⇌ N_{2} H_{4} (g) Δ H = 95 k J

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 :

For the gas phase reaction $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6},$ $Δ H$ = -32.7kcal carried out in a vessel, the equilibrium concentration of $C_{2} H_{4}$ can be increased by -

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:

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How can be the concentration of hydrazine, N2H4, could be increased in an equilibrium described by the equation.N2(g)+2H2(g)⇌N2H4(g) ΔH=95kJ

How can be the concentration of hydrazine, $N_{2} H_{4}$ , could be increased in an equilibrium described by the equation.

$N_{2} (g) + 2 H_{2} (g) ⇌ N_{2} H_{4} (g)$ $Δ H = 95 k J$