# chem- i reallyneed help

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Consider the following equations.
N2H4(l) + O2(g) N2(g) + 2 H2O(l) ÄH = -622.2 kJ
H2(g) + 1/2 O2(g) H2O(l) ÄH = -258.5 kJ
H2(g) + O2(g) H2O2(l) ÄH = -187.8 kJ

Use this information to calculate the enthalpy change for the reaction shown below.

N2H4(l) + 2 H2O2(l) N2(g) + 4 H2O(l) ÄH = ?

i really do0nt know how to do these

• chem- i reallyneed help -

Add equn 1 as is, to 2x equation 2, to 2x the reverse of equation 3 to obtain the equation you want. Then add the delta H values. When you multiply an equation you must multiply delta H values too. When reversing an equation, change the sign of delta H.

• chem- i reallyneed help -

so is it - 764.2?

• chem- i reallyneed help -

oops that's me ^

• chem- i reallyneed help -

763.6 is what I get.

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