Question

The equilibrium constant of the reaction

A₂ (g) + B₂ (g)$\rightleftarrows$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 100ºC, the amount of AB produced at equilibrium at 100ºC would be

• A. 0.93 mol
• B. 1.87 mol
• C. 2.80 mol
• D. 3.74 mol

Answer 1

Answer: (B)

1.87 mol

Answer 2

A₂ (g) + B₂ (g) $\rightleftarrows$ 2AB (g)

Initially: 1/3 (M) $\frac { 2 } { 3 }$(M)

At equilibrium: $\left( \frac { 1 } { 3 } - x \right)$M (M) 2x (M)

K_eq = 50 =

or 50 = $\frac { 36 x ^ { 2 } } { ( 1 - 3 x ) ( 2 - 3 x ) }$

or 50 (9x² – 9x + 2) = 36 x²

or 450 x² – 450x + 100 = 36x²

or 414x² – 450x + 100 = 0

or x = $\frac { + 450 - \sqrt { ( 450 ) ^ { 2 } - 4 \times 414 \times 100 } } { 2 \times 414 }$

or x = $\frac { + 450 - \sqrt { 202500 - 165600 } } { 2 \times 414 }$

or x = $\frac { 450 - \sqrt { 36900 } } { 2 \times 414 }$=$\frac { 450 - 192.1 } { 2 \times 414 }$

= 0.31 (M)

\ moles of AB produced = 0.31 × 6 = 1.86