Question

Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g)

Use the following:
https://study.ashworthcollege.edu/access/content/group/1b1c18e8-dde7-4a9d-a60a-ae7e447f5550/chemexam1q20.png

Answer 1

A. –291.9 kJ
B. –279.9 k
C. +279.9 kJ

Answer 2

can u make Hess cycle?

Answer 3

I have no idea what that is .. /.\

Answer 4

@iPwnBunnies ? :)

Answer 5

i better go without hess cycle first

Answer 6

Yeah..i don't get anything like nothing in Chemistry.. /.\

Answer 7

look you have 3 reactions
this is the main one H2O(s) → H2(g) + 1/2O2(g)
H2(g) + 1/2O2(g) → H2O (l) -285.9
H2O(s) → H2O (l) +6

Answer 8

what is similar in H2O(s) → H2(g) + 1/2O2(g) and H2(g) + 1/2O2(g) → H2O (l) reactions?

Answer 9

H2(g)+1/202(g) ????

Answer 10

yup ) in first one we GET H2(g)+1/202(g)
in second one we get H2O(l) FROM H2(g)+1/202(g)

Answer 11

Okay so what next? :)

Answer 12

next look at firs reaction and last one

Answer 13

H2O(s) → H2(g) + 1/2O2(g) and H2O(s) → H2O (l)

Answer 14

Okay ...soo?

Answer 15

do u see similarity in these 2 reactions?

Answer 16

H2O(s)?

Answer 17

yup
so in first reaction H2O(s) breaks down H2(g) + 1/2O2(g)
in second it changes to H2O(l)

Answer 18

now i'll connect all equations

Answer 19

H2O(s) → H2(g) + 1/2O2(g)
H2(g) + 1/2O2(g) → H2O (l)

we said that in first we get H2(g) + 1/2O2(g) and in second we get something From it right?

so if i were to connect these 2 reactions i could write it like this

H2O(s) → H2(g) + 1/2O2(g) → H2O (l)

do u agree?

Answer 20

Yeah :)

Answer 21

good now last reaction is H2O(s) → H2O (l)

Answer 22

do u see them in the equation i made?

Answer 23

they are in the corners right?

Answer 24

Yes

Answer 25

Yeah i know :)

Answer 26

so i can show it this way

Answer 27

so now i can make the Hess cycle

Answer 28

Okay:)