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# Enthalpy change watch

1. Can someone help me with this question:

Hydrazine may be obtained from the reaction between ammonia and hydrogen peroxide.
2NH3(g) + H2O2(l) → N2H4(l) + 2H2O(l) ∆rHo = −241.0 kJ mol−1
Work out the standard enthalpy change for the decomposition of hydrazine to its elements.
The standard enthalpy changes of formation in kJ mol−1 are:
NH3: −46.1; H2O2: −187.8; H2O: −285.8

Is 241 from reactants to products?
Second bit -92.2 + -187.8? = -280
Am I doing it right? Thanks in advance
2. (Original post by Super199)
Can someone help me with this question:

Hydrazine may be obtained from the reaction between ammonia and hydrogen peroxide.
2NH3(g) + H2O2(l) → N2H4(l) + 2H2O(l) ∆rHo = −241.0 kJ mol−1
Work out the standard enthalpy change for the decomposition of hydrazine to its elements.
The standard enthalpy changes of formation in kJ mol−1 are:
NH3: −46.1; H2O2: −187.8; H2O: −285.8

Is 241 from reactants to products?
Second bit -92.2 + -187.8? = -280
Am I doing it right? Thanks in advance
The equation enthalpy can be worked out from enthalpy of formation of products - enthalpy of formation of reactants

[ΔHf(N2H4) + 2 * ΔHf(H2O)] - [2*ΔHf(NH3) + ΔHf(H2O2)] = -241kJ

Substitute values and rearrange to find hydrazine
3. (Original post by charco)
The equation enthalpy can be worked out from enthalpy of formation of products - enthalpy of formation of reactants

[ΔHf(N2H4) + 2 * ΔHf(H2O)] - [2*ΔHf(NH3) + ΔHf(H2O2)] = -241kJ

Substitute values and rearrange to find hydrazine
Do I have to include all the negatives.
As what I have done is:
Hydrazine + 2(285.8)- (2x46.1 + 187.8)=241
Hydrazine + 291.6=241
Hydrazine = -50.6Kj mol-1
Does the negative -241 just show that it is an exothermic reaction?
4. (Original post by Super199)
Do I have to include all the negatives.
As what I have done is:
Hydrazine + 2(285.8)- (2x46.1 + 187.8)=241
Hydrazine + 291.6=241
Hydrazine = -50.6Kj mol-1
Does the negative -241 just show that it is an exothermic reaction?
you must include all the signs...

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