AB, $A_{2} and B_{2}$ are diatomic molecules, if the bond enthalpies of $A_{2}, AB and B_{2}$ are in the ratio of 1 : 1 : 0.5 and the enthalpy of formation of AB from $A_{2} and B_{2}$ is $- 100 kJ {mol}^{- 1}, combustion of A_{2} gives AO and ∆ H_{C} = - 1200 kJ {mol}^{- 1} .$ Bond energy of $(O = O) bond is 500 kJ {mol}^{- 1} .$ What is the bond enthalpy of (A - O) bond ?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$200 kJ {mol}^{- 1}$

$1650 kJ {mol}^{- 1}$

$400 kJ {mol}^{- 1}$

$1200 kJ {mol}^{- 1}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Bond enthalpy (BE) of $A_{2} = x, AB = x and B_{2} = \frac{x}{2}$

$\frac{1}{2} A_{2} + \frac{1}{2} B_{2} \overset{}{\to} AB - 100 kJ / {mol}^{- 1} . . . . . (i) \frac{1}{2} A_{2} + \frac{1}{2} O_{2} \overset{}{\to} AO - 1200 kJ / mol (given) A_{2} + O_{2} \overset{}{\to} 2 AO - 2400 kJ / mol . . . . (ii) O = O \overset{}{\to} 500 kJ / mol From eq . (i) \frac{1}{2} B . E_{A_{2}} + \frac{1}{2} B . E_{B_{2}} - B . E_{AB} = - 100 \frac{1}{2} x + \frac{x}{4} - x = - 100 - \frac{x}{4} = - 100 x = 400 kJ / mol Now, from eq . (ii) B . E_{A_{2}} + B . E_{O_{2}} - 2 B . E_{AO} = - 2400 x + (+ 500) - 2 B . E_{AO} = - 2400 400 + 500 - 2 B . E_{AO} = - 2400 - B . E_{A - O} = - 3300 B . E_{A - O} = 1650 kJ / mol .$