Consider the following data: f H ( N 2 H 4 , ? l ) = 50 kJ / mol , f H ( NH 3 , ? g ) = - 46 kJ / mol , B.E. ( N - H ) = 393 kJ / mol and B.E. ( H - H ) = 436 kJ / mol , vap H ( N 2 H 4 , l ) = 18 kJ / mol . The N - N bond energy i

Question

Consider the following data: $Δ_{f} H ° (N_{2} H_{4}, l) = 50 kJ / mol,$

$Δ_{f} H ° ({NH}_{3}, g) = - 46 kJ / mol, B.E. (N - H) = 393 kJ / mol$ and $B.E. (H - H) = 436 kJ / mol,$ $Δ_{vap} H (N_{2} H_{4}, l) = 18 kJ / mol .$

The $N - N$ bond energy in $N_{2} H_{4}$ is $(kJ / mol)$ .

Smart Achievers · Accepted Answer

(190)

$190 kJ/mol$

$\frac{1}{2} N_{2} + \frac{3}{2} H_{2} \to N H_{3},$

$Δ_{f} H ° ({NH}_{3}, g) = - 46 kJ / mol, B.E. (N - H) = 393 kJ / mol$ $= \frac{x}{2} + \frac{3}{2} \times 436 - 3 \times 393 = 958$

$x = 958$

$N_{2} H_{4} \to N_{2} + 2 H_{2}$

$Δ H = [Δ H_{vap} (N_{2} H_{4}) + 4 B.E. (N - H) + B.E. (N - N)] - [B.E. (N \equiv N) + 2 B.E. (H_{B . E .})]$

Solution

Q.
Consider the following data: $Δ_{f} H ° (N_{2} H_{4}, l) = 50 kJ / mol,$

$Δ_{f} H ° ({NH}_{3}, g) = - 46 kJ / mol, B.E. (N - H) = 393 kJ / mol$ and $B.E. (H - H) = 436 kJ / mol,$ $Δ_{vap} H (N_{2} H_{4}, l) = 18 kJ / mol .$

The $N - N$ bond energy in $N_{2} H_{4}$ is $(kJ / mol)$ .

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Solution

Q. Consider the following data: ΔfH°(N2H4, l)=50 kJ/mol, ΔfH°(NH3, g)=-46 kJ/mol, B.E.(N-H)=393 kJ/mol and B.E.(H-H)=436 kJ/mol, ΔvapH(N2H4, l)=18 kJ/mol. The N-N bond energy in N2H4 is (kJ/mol).

Ans. (190) 190 kJ/mol 12N2+32H2→NH3, Let B.E.(N≡N)=x-46=x2+32×436-3×393=958 x=958 N2H4→N2+2H2 ΔH=[ΔHvap(N2H4)+4 B.E.(N-H)+B.E.(N-N)]-[B.E.(N≡N)+2 B.E.(HB.E.)]

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Q.
Consider the following data: $Δ_{f} H ° (N_{2} H_{4}, l) = 50 kJ / mol,$

$Δ_{f} H ° ({NH}_{3}, g) = - 46 kJ / mol, B.E. (N - H) = 393 kJ / mol$ and $B.E. (H - H) = 436 kJ / mol,$ $Δ_{vap} H (N_{2} H_{4}, l) = 18 kJ / mol .$

The $N - N$ bond energy in $N_{2} H_{4}$ is $(kJ / mol)$ .