MyQuestionIcon

1

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

Enthalpy

Heat of atomi...

Question

Heat of atomisation of $N H_{3}$ and $N_{2} H_{4}$ are $x k c a l m o l^{- 1}$ and $y k c a l m o l^{- 1}$ respectively. Average bond energy of $N - - N$ bond is:

A

\frac{3 y - 4 x}{3} k c a l m o l^{- 1}

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

B

\frac{3 y - x}{5} k c a l m o l^{- 1}

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

C

\frac{2 y - 4 x}{4} k c a l m o l^{- 1}

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

D

None of these

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

Open in App

Solution

The correct option is A $\frac{3 y - 4 x}{3} k c a l m o l^{- 1}$
$N H_{3} ⟶ N (g) + 3 H (g); Δ H_{1} = x k c a l m o l^{- 1}$
$N_{2} H_{4} ⟶ 2 N (g) + 4 H (g); Δ H_{2} = y k c a l m o l^{- 1}$
$Δ H_{1} = 3 \times e_{N - - H} = x$ ...(i)
$Δ H_{2} = 4 \times e_{N - - H} + e_{N - - N} = y$ ...(ii)
From eqs. (i) and (ii)
$∴ y = 4. \frac{x}{3} + e_{N - - N}$
$∴ e_{N - - N} = y - \frac{4 x}{3}$
$= \frac{3 y - 4 x}{3} k c a l m o l^{- 1}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Heat of atomisation of

N H_{3}

N_{2} H_{4}

x k c a l m o l^{- 1}

y k c a l m o l^{- 1}

respectively. What will be average bond energy of

N - N

?

Q. Heat of atomisation of

N H_{3}

N_{2} H_{4}

x k c a l m o l^{- 1}

y k c a l m o l^{- 1}

respectively. What will be average bond energy of

N - N

?

Q. The standard heat of combustion of ethene gas is

- 330 K c a l m o l^{- 1}

C = C

bond energy (in kcal/mol) assuming that bond energy of

C - H

93.6 K c a l m o l^{- 1}

Δ H_{f}^{0}

C O_{2} (g)

H_{2} O (l)

- 94.2

- 61 K c a l m o l^{- 1}

respectively. Heat of atomisation of carbon and hydrogen are

150

51.5 K c a l m o l^{- 1}

$I f 2 x^{2} - 3 x y + y^{2} + x + 2 y - 8 = 0 t h e n \frac{d y}{d x} =$

Q. For the reaction

N_{2} + 3 X_{2} \to 2 N X_{3}

X = F, C l

(the average bond energies are

F - F = 155 k J m o l^{- 1}

N - F = 272 k J m o l^{- 1}

C l - C l = 242 k J m o l^{- 1}

N - C l = 200 k J m o l^{- 1}

N \equiv N = 941 k J m o l^{- 1}

). The heats of formation of

N F_{3}

N C l_{3}

k J m o l^{- 1}

, respectively, are closest to:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Enthalpy

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon