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

Question

If the bond dissociation energies of XY,X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and Δ Hf for the formation of XY is -200 text kJ text mol-1, the bond dissociation energy of X2 will be . (A)  400 text kJ text mol-1  (B)  300 text kJ text mol-1  (C)  200 text kJ text mol-1  (D)  800 text kJ text mol-1

Tardigrade Admin · Accepted Answer

X2+Y2 xrightarrow[]2XY Δ H=(BE)X-X+(BE)Y-Y-2(BE)X-Y If BE of (XY)=a Then BE of (XX)=a and BE of (Y-Y)=(a/2) Δ Hf=(X-Y)=-200 kJ ∴ -400 (for 2 moles XY ) =a+(a/2)-2a -400=-(a/2) or a=+800 text kJ

Q. 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 $Δ H_{f}$ for the formation of XY is $- 200 k J m o l^{- 1},$ the bond dissociation energy of $X_{2}$ will be .

$400 k J m o l^{- 1}$

$300 k J m o l^{- 1}$

$200 k J m o l^{- 1}$

$800 k J m o l^{- 1}$

$X_{2} + Y_{2} 2 X Y$
$Δ H = (BE)_{X - X} + (BE)_{Y - Y} - 2 (BE)_{X - Y}$
If BE of $(X Y) = a$
Then BE of $(XX) = a$
and BE of $(Y - Y) = \frac{a}{2}$
$Δ H_{f} = (X - Y) = - 200 k J$
$∴$ $- 400$ (for 2 moles $X Y$ )
$= a + \frac{a}{2} - 2 a$
$- 400 = - \frac{a}{2}$
or $a = + 800 k J$

Q. If the bond dissociation energies of XY,X2​ and Y2​ (all diatomic molecules) are in the ratio of 1:1:0.5 and ΔHf​ for the formation of XY is −200kJmol−1, the bond dissociation energy of X2​ will be .

400kJmol−1

300kJmol−1

200kJmol−1

800kJmol−1