Consider the reaction-4 NO 2 g - O 2 g - 2 N 2 O 5 g - - r H -111 kJ - If N 2 O 5 s is formed instead of N 2 O 5 g in the above reaction- the - r H value willbe-given- - H of sublimation for N 2 O 5 is 54 kJ mol -1 A- -219 kJB- -219 JC- -165 kJD- -54 kJ

The correct option is C −165 kJ Δ_r H = 111 kJ 111 54 = Δ H' Δ H' = 165 kJ ...

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Standard XII

Chemistry

Enthalpy

Consider the ...

Question

Consider the reaction:

$4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), Δ_{r} H = - 111 k J$ . If $N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $Δ_{r} H$ value will be:

(given, $Δ H$ of sublimation for $N_{2} O_{5}$ is $54 k J m o l^{- 1}$ )

+ 54 kJ

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+ 219 kJ

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− 219 J

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−165 kJ

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Solution

The correct option is D
−165 kJ

$Δ_{r} H = - 111 k J$

$- 111 - 54 = Δ H^{'}$

$Δ H^{'} = - 165 k J$

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Similar questions

Consider the reaction:

$4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), Δ_{r} H = - 111 k J$ . If $N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $Δ_{r} H$ value will be:

(given, $Δ H$ of sublimation for $N_{2} O_{5}$ is $54 k J m o l^{- 1}$ )

Q. Consider the reaction,

4 {N O}_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), Δ_{r} H = - 111 k J .

N_{2} O_{5} (s)

is formed instead of

N_{2} O_{5} (g)

in the above reaction, the

Δ_{r} H

value will be
( Given,

Δ H

of sublimation for

N_{2} O_{5}

54 k J {m o l}^{- 1}

)

Q. In the reaction

4 N O_{2_{(g)}} + O_{2}_{(g)} \to 2 N_{2} O_{5_{(g)}}, Δ H = - 111

kJ if

N_{2} O_{5_{(s)}}

is formed instead of

N_{2} O_{5_{(g)}}

, the

Δ

H value in kJ is:

[Δ H_{s u b l i m a t i o n}

for

N_{2} O_{5} = 54

m o l^{- 1}]

Q. For a first order reaction involving decomposition of

N_{2} O_{5}

the following information is available

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

rate

= k [N_{2} O_{5}]

N_{2} O_{5} (g) \to 2 N O_{2} (g) + 1 / 2 O_{2} (g)

rate

= k^{'} [N_{2} O_{5}]

Q. For the reaction

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

- \frac{Δ [N_{2} O_{5}]}{Δ t} = k_{1} [N_{2} O_{5}], + \frac{[N O_{2}]}{Δ t} = k_{2} [N_{2} O_{5}], + \frac{Δ [O_{2}]}{Δ t} = k_{3} [N_{2} O_{5}]

Thus,

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Consider the reaction: 4NO2(g) + O2(g) → 2N2O5(g), ΔrH = −111kJ. If N2O5(s) is formed instead of N2O5(g) in the above reaction, the ΔrH value will be: (given, ΔH of sublimation for N2O5 is 54 kJ mol−1)

The correct option is D −165 kJ ΔrH = −111 kJ −111 − 54 = ΔH′ ΔH′ = −165 kJ

Consider the reaction:

$4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), Δ_{r} H = - 111 k J$ . If $N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $Δ_{r} H$ value will be:

(given, $Δ H$ of sublimation for $N_{2} O_{5}$ is $54 k J m o l^{- 1}$ )

The correct option is D
−165 kJ

$Δ_{r} H = - 111 k J$

$- 111 - 54 = Δ H^{'}$

$Δ H^{'} = - 165 k J$