Consider the following reversible reaction- In a 3-00 litre container- the following amounts are found in equilibrium at 400- C -0-04 moleN 2- 0-5 moleH 2 and 0-04 moleN H 3 - Evaluate K c - N 2 g -3 H 2 g - 2 N H 3 g A- 2-88B- 1-88C- 2D- 0-84

The correct option is A 2.88→ N_2(g) + 3H_2(g) ⇌ 2NH_3(g) 0.04 mole 0.5 mole 0.04 mole → Here V = 3.00 L K_c = [NH_3]^2[N_2][H_2]^3 = ( 0.043 )^2 ( 0.043 ) × ...

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Byju's Answer

Standard XII

Chemistry

Characteristics of Chemical Equilibrium

Consider the ...

Question

Consider the following reversible reaction. In a 3.00 litre container, the following amounts are found in equilibrium at $400^{\circ} C : 0.04 mole N_{2}, 0.5 mole H_{2}$ and $0.04 mole N H_{3}$ . Evaluate $K_{c}$ . $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

2.88

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Solution

The correct option is A 2.88
$\to N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
0.04 mole 0.5 mole 0.04 mole
$\to$ Here V = 3.00 L
$K_{c} = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}$
$= \frac{{(\frac{0.04}{3})}^{2}}{(\frac{0.04}{3}) \times {(\frac{0.5}{3})}^{3}} (Concentraction = \frac{mole}{Volume})$
$= 2.88$

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Similar questions

Q. The unit of

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

Q. In the reversible reaction

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

when 1 mol of

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Q. When

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Q. The equilibrium constant

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P_{4} (g) ⇌ 2 P_{2} (g)

1.4

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1.4

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Suppose that

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Consider the following reversible reaction. In a 3.00 litre container, the following amounts are found in equilibrium at 400∘C:0.04moleN2,0.5moleH2 and 0.04moleNH3. Evaluate Kc.N2(g)+3H2(g)⇌2NH3(g)

The correct option is A 2.88→N2(g)+3H2(g)⇌2NH3(g) 0.04 mole 0.5 mole 0.04 mole → Here V = 3.00 L Kc=[NH3]2[N2][H2]3 =(0.043)2(0.043)×(0.53)3(Concentraction=moleVolume) =2.88

Consider the following reversible reaction. In a 3.00 litre container, the following amounts are found in equilibrium at $400^{\circ} C : 0.04 mole N_{2}, 0.5 mole H_{2}$ and $0.04 mole N H_{3}$ . Evaluate $K_{c}$ . $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$