• anonymous
Given: NH3--> 1/2N2H4+1/2H2 A) N2+2H2-->N2H4 deltaH = 95.4kJ B) N2+3H2-->2NH3 deltaH = -92.2 Can you show me the steps to calculate this?
  • Stacey Warren - Expert
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  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
If you google "Chemistry Equation Balancer" you'll get the calculator that does this FOR you (;
  • raspberryjam
Okay so you have the equation: NH3 --> 1/2N2H4 + 1/2H2 Notice in the equation you only have 1 NH3. Now look at equation B...notice anything? Try flipping the equation around (so that NH3 is on the reactant side) and dividing that equation in half. Remember to make the delta H positive and divide the delta H in half too. So now for equation B you should have: NH3 -- > 1/2N2 + 3/2H2 Now try cutting equation A in half since you only need 1/2N2H4. Remember to divide delta H in half too. So for equation A you should have: 1/2N2 + H2 --> 1/2N2H4 Now add A and B together... NH3 + 1/2N2 + H2 --> 1/2N2 + 3/2H2 + 1/2N2H4 and you should be able to cancel out some compounds: NH3 --> 1/2H2 + 1/2N2H4 aka what you were given! Now in terms of math what we did was flipped equation B and divided the equation in half. Therefore we need to flip the sign for delta H of B (from negative to positive) and divide delta H in half. Then we divided equation A in half so we should also divide the delta H for A in half. Then just add the delta Hs for A and B together to get the total enthalpy.

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