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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

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False

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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
${Δ H}_{r} = X_{1} + {3 X}_{2} - {2 \times 3 X}_{3}$
${Δ H}_{r} = X_{1} + {3 X}_{2} - {6 X}_{3}$

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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

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and

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and

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;

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

and

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= 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

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and

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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
${Δ H}_{r} = X_{1} + {3 X}_{2} - {2 \times 3 X}_{3}$
${Δ H}_{r} = X_{1} + {3 X}_{2} - {6 X}_{3}$