Given the reaction 2 Na2 02 (s) + 2 H2o(i) —- 4 NaOH(s) + 02(g) triangle H= -109 KJ

determine the following reactions.
16 Na2o2(s)+ 16 H2o(i)———- 32 NaOH(s) + 8 O2(g)
2 NaOH(s)—-+ 1/2 o2(g)——-Nao2(s)+ H2o(I)

You really need to be more careful with your typing. I have an excuse when I make typos and omit characters; I'm old and can't see very well. You don't have that excuse. You're in a hurry. You can't hurry your way through chemistry. :-).

Equation 1 has dH = -109 kJ.
Look at equation 2. That is just 8x the first one; therefore dH must be 8*-109 kJ.

Look at equation 3. That one is 1/2 of the reverse of equation 1; therefore, dH is (1/2*-109 kJ) then change the sign since the equation is reversed. Note you made a typo here. I think you meant Na2O2(s)

My apologies. I will type it over.

Given the reaction 2 Na2 02 (s) + 2 H2o(i) —- 4 NaOH(s) + 02(g)

I couldn't make out the symbol however its triangle H= -109 KJ

determine the following reactions.
1. 16 Na2o2(s)+ 16 H2o(i)———- 32 NaOH(s) + 8 O2(g)
2. 2 NaOH(s)———— + 1/2 o2(g)——-Nao2(s)+ H2o(I)

The triangle is the Greek for delta.

Ok. I retyped the question straight from the book.hopefully its correct now.

Thanks. My answer stays the same. But I still think that last NaO2 is Na2O2. NaO2 doesn't make sense with the rest of the problem.

To determine the enthalpy change (∆H) for the given reactions, we can use the concept of Hess's law. According to Hess's law, the total enthalpy change for a reaction is independent of the pathway taken to reach the final products. This allows us to manipulate equations and combine them to obtain new reactions.

Let's start with the given reaction:

Reaction 1: 2 Na2O2(s) + 2 H2O(l) → 4 NaOH(s) + O2(g) ∆H = -109 kJ

Now, let's determine the enthalpy change for the first reaction you provided:

Reaction 2: 16 Na2O2(s) + 16 H2O(l) → 32 NaOH(s) + 8 O2(g)

To obtain the enthalpy change for this reaction, we need to multiply the given reaction by a factor so that the coefficients match those in reaction 2. Since the coefficients for the given reaction are already balanced, we don't need to change them. However, the enthalpy change needs to be adjusted accordingly. Multiplying the given reaction by 8 will ensure that the number of moles of products and reactants match reaction 2:

8 × (Reaction 1):
16 Na2O2(s) + 16 H2O(l) → 32 NaOH(s) + 8 O2(g) ∆H = -109 kJ × 8 = -872 kJ

Hence, the enthalpy change for the reaction "16 Na2O2(s) + 16 H2O(l) → 32 NaOH(s) + 8 O2(g)" is -872 kJ.

Moving on to the second reaction you provided:

Reaction 3: 2 NaOH(s) + 1/2 O2(g) → NaO2(s) + H2O(l)

To determine the enthalpy change for this reaction, we can use reaction 1 as a reference. By flipping reaction 1 and changing its sign, we can obtain reaction 3:

-0.5 × (Reaction 1):
-1/2 (2 Na2O2(s) + 2 H2O(l) → 4 NaOH(s) + O2(g)) ∆H = 0.5 × 109 kJ = 54.5 kJ

So, the enthalpy change for the reaction "2 NaOH(s) + 1/2 O2(g) → NaO2(s) + H2O(l)" is 54.5 kJ.

Therefore, the enthalpy changes for the provided reactions are as follows:

Reaction 1: 2 Na2O2(s) + 2 H2O(l) → 4 NaOH(s) + O2(g) ∆H = -109 kJ

Reaction 2: 16 Na2O2(s) + 16 H2O(l) → 32 NaOH(s) + 8 O2(g) ∆H = -872 kJ

Reaction 3: 2 NaOH(s) + 1/2 O2(g) → NaO2(s) + H2O(l) ∆H = 54.5 kJ