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The formation of H(2)O(2) in the upper a...

The formation of `H_(2)O_(2)` in the upper atmosphere follows the mechanism
`H_(2)O+Orarr2OH rarrH_(2)O_(2)`
`DeltaH=72 " kJ" " mol"^(-1), E_(a)=77 " kJ mol"^(-1)`
`E_(a)` for the backward process is

A

149 kJ `"mol"^(-1)`

B

`-149 " kJ mol"^(-1)`

C

5 kJ `"mol"^(-1)`

D

`-5` kJ `"mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C

`E_(a)("backward") = (77-72)kJ mol^(-1)`
`=5 kJ mol^(-1)`
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There is formation of H_(2)O_(2) in the upper atomsphere. H_(2)O + O rarr 2OH rarr H_(2)O_(2) Delta H = 72 kJ mol^(-1) , E_(a) = 77 kJ mol^(-1) What will be the E_(a) for bimolecular recombination of two OH radicals to form H_(2)O and O in kJ.

In gaseous reaction, important for the understanding of the upper atmosphere H_(2)O and O react bimolecularly to from two OH readicals. DeltaH for this reaction is 72 kJ at 500 K and E_(a) is 77 kJ "mol"^(-1) , then E_(a) for the bimolecular recombination of two OH readicals to form H_(2)O and O is :

Calculate the enthalpy of formation of sucrose (C_11H_22O_11) from the following data: (i) C_(12)H_(22)O_(11) + 12O_(2) rarr 12CO_(2) + 11 H_(2)O, DeltaH = -5200.7 kJ mol^(-1) (ii) C + O_(2) rarr CO_(2), DeltaH = -394.5 kJ mol^(-1) (iii) H_(2) + 1/2 O_(2) rarr H_(2)O, DeltaH = H_(2)O, DeltaH = -285.8 kJ mol^(-1) .

The average O-H Bond energy in H_(2)O with the help of following data : (P) H_(2)O(l)toH_(2)O(g)," "DeltaH=+40.6 kJ mol^(-1) (Q) 2H(g)toH_(2)(g)," "DeltaH=-435.0 kJ mol^(-1) (R ) O_(2)(g)to2O(g)," "DeltaH=+489.6 kJ mol^(-1) (S) 2H_(2)(g)+O_(2)(g)to2H_(2)O(l), " "DeltaH=-571.6 kJ mol^(-1)

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