Easy Chem Question; Please help!!! Given the following data:

Question

Given the following data,

H2(g) 1/2 O2(g) => H2O(g) DGo = -242.4

2 H(g) O(g) => H2O(g) DGo = -927.8

2 H(g) 2 O(g) => H2O2(g) DGo = -1,068.5

2 O(g) => O2(g) DGo = -493.3

H2O2(l) => H2O2(g) DGo = 44.7

Determine DGfo for H2O2(l)

Answer 1

(A) H2(g) 1/2 O2(g) => H2O(g) DGo = -242.4

(B) 2 H(g) O(g) => H2O(g) DGo = -927.8

(C) 2 H(g) 2 O(g) => H2O2(g) DGo = -1,068.5

(D) 2 O(g) => O2(g) DGo = -493.3

(E) H2O2(l) => H2O2(g) DGo = 44.7

=================

find DGfo for H2O2(l) :

H2 & O2 --> H2O2 (l)

only the reverse of (E) will give you H2O2(l) as a product:

- (E):   H2O2(g) => H2O2(l) ..... DGo = - 44.7

only by combining that with  (C) can you cancel out the H2O2(g):

(C) 2 H(g) 2 O(g) => H2O2(g) ....DGo = -1,068.5

only by combining with the reverse of (B) can you cancel out the 2H:

- (B)  H2O(g) => 2 H(g) O(g)  ....DGo = + 927.8

only by combining with 1/2 of the reverse of (D) can you cancel out the last O:

- 1/2 (D)   1/2 O2(g) => 1 O(g)   .... DGo = + 246.65

only by combining with (A) can you cancel out the water:

(A) H2(g) 1/2 O2(g) => H2O(g) DGo = -242.4

your job is to start crossing off every thing that identical amounts of reactants  & products, showing that the only thing left is what you wanted, a dHf reaction for H2O2(l) .....

which is : H2 & O2 --> H2O2 (l)

when I reversed the equation I reversed the sign of its dG, when I had to 1/2 (D), I 1/2 'ed its dG

hesses law says if however you combine equations gives you want you wanted , then do the same with their dG's & you will get the dG you wanted:

dGo = - 44.7  -1,068.5  + 927.8 + 246.65  -242.4

your answer is: -181.2 kJ

how do you define "easy"