The formation of H2O2 in the upper atmosphere follows the mechanism- H2O-O- 2OH - H2O2- - H-72 kJ mol-1- Ea-77 kJ mol-1- Then- Ea for the backward reaction will be per mol-

The correct option is D +5 kJRelation used; #160; Δ H=E_a(for.) E_a(back) or E_a(back)=E_a(for) Δ H E_a(b)=77 72=+5 kJ . ...

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Standard XII

Chemistry

Activation Energy

The formation...

Question

The formation of $H_{2} O_{2}$ in the upper atmosphere follows the mechanism: $H_{2} O + O \to 2 O H \to H_{2} O_{2}$ ; $Δ H = 72 k J m o l^{- 1}$ , $E_{a} = 77 k J$ $m o l^{- 1}$ .

Then, $E_{a}$ for the backward reaction will be (per mol):

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Solution

The correct option is D $+ 5$ kJ
Relation used;

$Δ H = E_{a (f o r .)} - E_{a (b a c k)}$

or $E_{a (b a c k)} = E_{a (f o r)} - Δ H$

$E_{a (b)} = 77 - 72 = + 5 k J$ .

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Similar questions

Q. In gaseous reaction important for the understanding of the upper atmosphere

H_{2} O

and

O

react bimolecularly to form two

O H

radicals.

Δ H

for this reaction is 72 kJ at 500 K and

E_{a}

is 77 kJ

{m o l}^{- 1}

E_{a}

for the bimolecular recombination of two OH radicals to form

H_{2} O

and

O

is:

(in kJ

{m o l}^{- 1}

)

I

E_{a} = 15

k J

{m o l}^{- 1};

Δ H = - 70

k J

{m o l}^{- 1}

I I

E_{a} = 30

k J

{m o l}^{- 1};

Δ H = - 15

k J

{m o l}^{- 1}

I I I

E_{a} = 60

k J

{m o l}^{- 1};

Δ H = + 20

k J

{m o l}^{- 1}

If above reactions are at same frequency factor and temperature, then fastest and slowest reactions are:

Q. In gaseous reactions, important for the understanding of the upper atmosphere

H_{2} O

and

O

react bimolecularly to form two

O H

radicals.

Δ H

for this reaction is

72

k J

500

K

and

E_{a}

77

k J

{m o l}^{- 1}

; then

E_{a}

for the bimolecular recombination of two

O H

radicals to form

H_{2} O

and

O

is:

Q. In gaseous reactions important for understanding the upper atmosphere,

H_{2} O

and O react biomolecularly to form two OH radicals.

Δ H

for this reaction is 72 kJ at 500 K and

E_{a} = 77 k J m o l^{- 1}

, then calculate

E_{a}

for the biomolecular recombination of

2 O H

radicals to form

H_{2} O

& O at 500 K.

Q. Heat of formation of

H_{2} O

- 188

kJ/mol and

H_{2} O_{2}

- 286

kJ/mol. The enthalpy change for the following reaction is:

2 H_{2} O_{2} \to 2 H_{2} O + O_{2}

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The formation of H2O2 in the upper atmosphere follows the mechanism: H2O+O→2OH→H2O2; ΔH=72 kJmol−1, Ea=77 kJ mol−1. Then, Ea for the backward reaction will be (per mol):

The correct option is D +5 kJRelation used; ΔH=Ea(for.)−Ea(back)or Ea(back)=Ea(for)−ΔHEa(b)=77−72=+5 kJ.

The formation of $H_{2} O_{2}$ in the upper atmosphere follows the mechanism: $H_{2} O + O \to 2 O H \to H_{2} O_{2}$ ; $Δ H = 72 k J m o l^{- 1}$ , $E_{a} = 77 k J$ $m o l^{- 1}$ .

Then, $E_{a}$ for the backward reaction will be (per mol):

The correct option is D $+ 5$ kJ
Relation used;

$Δ H = E_{a (f o r .)} - E_{a (b a c k)}$

or $E_{a (b a c k)} = E_{a (f o r)} - Δ H$

$E_{a (b)} = 77 - 72 = + 5 k J$ .