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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

If the bond dissociation energies of $XY, X_2 \: and \: Y_2$(all diatomic molecules) are in the ratio of 1:1:0.5 and $ \Delta H_f $ for the formation of XY is -200 kJ/mol. The bond dissociation energy of $X_2$ will be

$\begin {array} {1 1} (A)\;400\: kJ/mol & \quad (B)\;300\: kJ/mol \\ (C)\;200\: kJ/mol & \quad (D)\;None\: of \: these \end {array}$

1 Answer

$X_2 + Y_2 \rightarrow 2XY$
$ \Delta H = (BE)X-X + (BE)Y-Y – 2(BE)X-Y$
If $(BE)$ of $X-Y = a$
then $(BE) $ of $(X-X) = a$
and $(BE) $ of $(Y-Y) = \large\frac{ a}{2}$
So, $ \Delta Hf(X-Y) = -200\: kJ$
So, $-400 $ (for 2 moles $XY) = a + \large\frac{a}{2}$$ – 2a$
$ \Rightarrow -400 = \large\frac{-a}{2}$
$ \Rightarrow a = 800 \: kJ$
The bond dissociation energy of $X_2 = 800\: kJ/mol$
Ans : (D)
answered Mar 18, 2014 by thanvigandhi_1
 

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