464

You visited us 464 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Hydrogenation of Alkynes

If BE bond en...

Question

If BE (bond energy) of N $\equiv$ N bond is $X_{1}$ that of H H bond is $X_{2}$ and N H bond is $X_{1}$ then enthalpy change of the reaction is
$N_{2} + {3 H}_{2} ⟶ {2 N H}_{3}$
${Δ H}_{r} = X_{1} + {3 X}_{2} - {2 X}_{3}$

True

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

False

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

Open in App

Solution

The correct option is B False
As we know,
Heat of reaction = bond energy of reactant - bond energy of product
$N_{2} + {3 H}_{2} ⟶ {2 N H}_{3}$
so
${Δ H}_{r} = X_{1} + {3 X}_{2} - {2 \times 3 X}_{3}$
${Δ H}_{r} = X_{1} + {3 X}_{2} - {6 X}_{3}$

Suggest Corrections

Similar questions

If bond energy of $N - H, H - H$ and $N \equiv N$ are a, b and c respectively, then $Δ H$ for the reaction, $2 N H_{3} \to N_{2} + 3 H_{2}$ is?

Q. The bond energies of

N \equiv N

H - H

and

N - H

bonds are

945,

436

and

391

KJ/mole respectively. The enthalpy of the reaction

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

Q. Calculate the enthalpy of formation of ammonia from the following bond energy data:

(N - N)

bond

= 398 k J {m o l}^{- 1}

;

(H - H)

bond

= 435 k J {m o l}^{- 1}

and

(N \equiv N)

bond

= 945.36 k J {m o l}^{- 1}

Given the bond energies $N \equiv N$ , $H - H$ and $N - H$ bonds are $945, 436$ and $391 k J m o l e^{- 1}$ respectively, the enthalpy of the following reaction $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ is

Q. If

x_{1}

x_{2}

and

x_{3}

are enthalpies

H - H, O = O

and

O - H

bonds respectively, and

x_{4}

is the enthalpy of vaporization of water. Estimate the standard enthalpy of combustion of hydrogen.

Join BYJU'S Learning Program

If BE (bond energy) of N ≡ N bond is X1 that of H H bond is X2 and N H bond is X1 then enthalpy change of the reaction isN2+3H2⟶2NH3ΔHr=X1+3X2−2X3

The correct option is B FalseAs we know,Heat of reaction = bond energy of reactant - bond energy of productN2+3H2⟶2NH3soΔHr=X1+3X2−2×3X3ΔHr=X1+3X2−6X3

If BE (bond energy) of N $\equiv$ N bond is $X_{1}$ that of H H bond is $X_{2}$ and N H bond is $X_{1}$ then enthalpy change of the reaction is
$N_{2} + {3 H}_{2} ⟶ {2 N H}_{3}$
${Δ H}_{r} = X_{1} + {3 X}_{2} - {2 X}_{3}$

The correct option is B False
As we know,
Heat of reaction = bond energy of reactant - bond energy of product
$N_{2} + {3 H}_{2} ⟶ {2 N H}_{3}$
so
${Δ H}_{r} = X_{1} + {3 X}_{2} - {2 \times 3 X}_{3}$
${Δ H}_{r} = X_{1} + {3 X}_{2} - {6 X}_{3}$