Hess law of constant summation

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Enthalpy of formation of $A_{2} B$ is +1000 KJ. Ratio of bond enthalpies of $A_{2}, B_{2}$ and $A_{2} B$ is 1 : 2 : 1. Bond enthalpy of $B_{2}$ would be

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Ratio of bond enthalpies of $A_{2} : B_{2} : A_{2} B$ is 1 : 2 : 1
B.E of $A_{2} = x$
B.E of $B_{2} = 2 x$
B.E of $\frac{1}{2} B_{2} = x$
B.E of $A_{2} B = x$
$∴ \underset{x}{A_{2}} + \underset{x}{\frac{1}{2} B_{2}} \overset{}{\to} \underset{x}{A_{2} B}; ΔH = + 1000 kJ$
$ΔH (B . D . E) = H_{R} - H_{P}$
$+ 1000 kJ = 2 x - x$
$x = 1000 kJ$
B.D.E of $B_{2}$ = $2 x = + 2000 kJ$

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Hess law of constant summation

Remember concepts with our Masterclasses.

Enthalpy of formation of A2B is +1000 KJ. Ratio of bond enthalpies of A2,B2 and A2B is 1 : 2 : 1. Bond enthalpy of B2would be

Ratio of bond enthalpies of A2:B2:A2B is 1 : 2 : 1B.E of A2=xB.E of B2=2xB.E of 12B2=xB.E of A2B=x∴A2x+12B2x→ A2Bx;ΔH=+1000kJΔHB.D.E=HR−HP+1000kJ =2x -xx=1000kJB.D.E of B2 = 2x =+2000kJ

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Enthalpy of formation of $A_{2} B$ is +1000 KJ. Ratio of bond enthalpies of $A_{2}, B_{2}$ and $A_{2} B$ is 1 : 2 : 1. Bond enthalpy of $B_{2}$ would be