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Factors Affecting Alpha

The equilibri...

Question

The equilibrium constant of the reaction $A 2 (g) + B_{2} (g) ⇌ 2 A B (g)$ at 373 K is 50. If 1L of flask containing 1 mole of $A_{2}$ (g) is connected to 2L flask containing 2 moles
$B_{2} (g) a t 100^{o} - C$ , the amount of AB produced at equilibrium at $100^{o} - C$ would be

0.93 mol

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1.87 mol

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2.80 mol

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3.74 mol

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Solution

The correct option is B 1.87 mol
$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g) I n i t i a l l y : \frac{1}{3} (M) \frac{2}{3} (M) A t e q u i l i b r i u m : (\frac{1}{3} - x) M (\frac{2}{3} - x) (M) 2 x (M) K_{e q} = 50 = \frac{4 x^{2}}{(\frac{1}{3} - x) (\frac{2}{3} - x)} o r 50 = \frac{36 x^{2}}{(1 - 3 x) (2 - 3 x)} o r 50 (9 x^{2} - 9 x + 2) = 36 x^{2} o r 450 x^{2} - 450 x + 100 = 36 x^{2} o r 414 x^{2} - 450 x + 100 = 0 o r x = \frac{+ 450 - \sqrt{(450)^{2} - 4 \times 414 \times 100}}{2 \times 414} o r x = \frac{+ 450 - \sqrt{(202500 - 165600}}{2 \times 414} o r x = \frac{450 - \sqrt{36900}}{2 \times 414} = \frac{450 - 192.1}{2 \times 414} = 0.31 (M) ∴ m o l e s o f A B p r o d u c e d = 2 \times 0.31 \times 3 = 1.86$

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Similar questions

The equilibrium constant of the reaction $A 2 (g) + B 2 (g) ⇌ 2 A B (g)$ at $100^{\circ} C$ is 50.

If one litre flask containing one mole of $A_{2}$ is connected to a two litre flask containing two moles of $B_{2}$ the number of moles of AB formed at 373 K will be -

Q. The equilibrium constant of the reaction

A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)

100

50

. If a one litre flash containing one mole of

A_{2}

is connected to a two litre flask containing two moles of

B_{2}

how many moles will be formed at

373 K

Q. The equilibrium constant of the reaction,

A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)

100^{o} C

50

. If a one-liter flask containing

1 m o l e

A_{2}

is connected to a two-liter flask containing

2 m o l e s

B_{2}

, how many moles of

A B

will be formed at

373 K

Q. The equilibrium constant

K_{c}

of the reaction,

A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)

100^{\circ} C

64

. If a one litre flask containing one mole of

A_{2}

is connected to a two litre flask containing one mole of

B_{2},

how many moles of

A B

will be formed at

373 K

Q. The

K_{c}

for,

A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)

100

50

. If one litre flask containing one mole of

A_{2}

is connected with a two litre flask containing 2 mole of

B_{2}

, how many mole of

A B

will be formed at

100

, rounded to the nearest integer?

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The equilibrium constant of the reaction A2(g)+B2(g)⇌2AB(g) at 373 K is 50. If 1L of flask containing 1 mole of A2 (g) is connected to 2L flask containing 2 moles B2(g) at 100o−C, the amount of AB produced at equilibrium at 100o−C would be

The equilibrium constant of the reaction $A 2 (g) + B_{2} (g) ⇌ 2 A B (g)$ at 373 K is 50. If 1L of flask containing 1 mole of $A_{2}$ (g) is connected to 2L flask containing 2 moles
$B_{2} (g) a t 100^{o} - C$ , the amount of AB produced at equilibrium at $100^{o} - C$ would be