You visited us 1 times! Enjoying our articles? Unlock Full Access!

Heavy Water

Find the enth...

Question

Find the enthalpy of formation of Hydrogen flouride on the basis of following data :
Bond energy of $H - H = 434 k J m o l^{- 1}$
Bond energy of $F - F = 158 k J m o l^{- 1}$
Bond energy of $H - F = 565 k J m o l^{- 1}$

300 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

269 k J m o l^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

250 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

275 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $269 k J m o l^{- 1}$
We have to find enthalpy for the reaction
$\frac{1}{2} H_{2} (g) + \frac{1}{2} F_{2} (g) \to H F (g)$
$Δ H = B . E (R e a c t a n t s) - B . E (P r o d u c t s)$
$= \frac{1}{2} B . E (H_{2}) + \frac{1}{2} B . E (F_{2}) - B . E (H F)$
$= \frac{1}{2} \times 434 + \frac{1}{2} \times 158 - 565$
$= 269 k J m o l^{- 1}$

Suggest Corrections

Similar questions

H - H b o n d = 434 k J m o l^{- 1}

F - F b o n d = 158 k J m o l^{- 1}

H - F b o n d = 565 k J m o l - 1

E_{H - H} = 104.2 k c a l m o l^{- 1}

E_{F - F} = 36.6 k c a l m o l^{- 1}

E_{H - F} = 134.6 k c a l m o l^{- l}

X_{H} = 2.1.

E_{H - H} = 104.2 k c a l m o l^{- 1}

E_{F - F} = 36.6 k c a l m o l^{- 1}

E_{H - F} = 134.6 k c a l m o l^{- l}

X_{H} = 2.1.

E_{H - H} = 104.2 k c a l m o l^{- 1}

E_{C - C} = 83.1 k c a l m o l^{- 1}

E_{C - H} = 98.8 k c a l m o l^{- l}

X_{H} = 2.1.

H_{2} + F_{2} \to 2 H F

Join BYJU'S Learning Program