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 - 1 H for the formation of XY is -200 kJ mol -1- The bond dissociates energy of X 2 will be-

The correct option is D 800 kJ mol^ 1 X_2+Y_2→2XY Δ H=(BE)X_2+(BE)Y_2 2(BE)X Y If (BE) of X Y=a then (BE) of Y Y=a2 ...

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Standard XII

Physics

Calorimetry

If the bond d...

Question

If the bond dissociation energies of $X Y$ , $X_{2}$ and $Y_{2}$ (all diatomic molecules) are in the ratio of $1 : 1 : 0.5$ and $Δ_{1} H$ for the formation of XY is $- 200 k J {m o l}^{- 1}$ . The bond dissociates energy of $X_{2}$ will be:

100 k J m o l^{- 1}

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200 k J m o l^{- 1}

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800 k J m o l^{- 1}

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400 k J m o l^{- 1}

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Solution

The correct option is D $800 k J m o l^{- 1}$
$X_{2} + Y_{2} \to 2 X Y$
$Δ H = (B E) X_{2} + (B E) Y_{2} - 2 (B E) X Y$

If $(B E)$ of $X - Y = a$
then $(B E)$ of $Y - Y = \frac{a}{2}$
and $(B E)$ of $X - X = a$

So, $Δ H_{f} (X - Y) = - 200 k J$
So, $- 400 (f o r 2 m o l e s X Y) = a + \frac{a}{2} - 2 a$
$\Rightarrow - 400 = \frac{- a}{2}$
$a = 800 k J$

The bond dissociation energy of $X_{2} = 800 k J / m o l$

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If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and Δ1H for the formation of XY is −200 kJ mol−1. The bond dissociates energy of X2 will be:

The correct option is D 800 kJ mol−1X2+Y2→2XYΔH=(BE)X2+(BE)Y2−2(BE)XYIf (BE) of X−Y=athen (BE) of Y−Y=a2and (BE) of X−X=aSo, ΔHf(X−Y)=−200 kJSo, −400 (for 2 moles XY)=a+a2−2a⇒−400=−a2a=800 kJThe bond dissociation energy of X2=800 kJ/mol

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