Question

what is the standard enthalpy change for the following reaction Na2O(5) + H2O(1) → 2 NaOH(5)

Answer 1

Answer:

2 NaOH=−184

ΔNaOH=−96

2 naOH=ΔNa

2

O=−238

−184+238=ΔNa

2

O+54 KI=ΔNa

2

O

or

2 Na+2H

3

+2H

2

O→2 NaOH+H

2

N a

2

O

(s)

+H

2

O

(l)

→2 NaOH

(aq)

−286 H

2(g)

+

2

1

O

2(g)

→H

2

O

(l)

....(1)

2 Na

(3)

+M

2

O

(l)

→Na

2

O

(s)

+H

2(q)

...(2)

−184−286−230

(1)+(2)

2Na

(s)

+

2

1

O

2(g)

→Na

2

O

(s)

−54+−184+238+54 KI

Explanation:

BY TEJAS KULKARNI PLEASE FOLLOW AND MARK AS BRAINLIST