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
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.
so is it - 764.2?
oops that's me ^
763.6 is what I get.
To calculate the enthalpy change for the given reaction, you can use the concept of Hess's Law. Hess's Law states that if a reaction can be expressed as a series of intermediate reactions, then the overall enthalpy change is the sum of the enthalpy changes of those intermediate reactions.
In order to use Hess's Law, you need to understand that enthalpy change is a state function; this means that it depends only on the initial and final states of the reaction and is independent of the pathway taken. Therefore, you can rearrange the given equations and use them to construct the desired reaction. Here's how you can do it:
Step 1: Multiply the given equations to obtain the desired reaction.
Multiply equation 1 by 2 to cancel out the H2O(l) term:
2N2H4(l) + 4O2(g) → 2N2(g) + 4H2O(l) ΔH = -1244.4 kJ
Step 2: Multiply equation 2 by 2 to match the number of moles for H2O(l):
2H2(g) + O2(g) → 2H2O(l) ΔH = -517.0 kJ
Step 3: Multiply equation 3 by 2 to match the number of moles for H2O2(l):
2H2(g) + 2O2(g) → 2H2O2(l) ΔH = -375.6 kJ
Step 4: Add the equations obtained in steps 1, 2, and 3 to obtain the desired reaction:
2N2H4(l) + 4H2O2(l) → 2N2(g) + 8H2O(l) ΔH = -1244.4 kJ + (-375.6 kJ) + (-517.0 kJ)
Step 5: Calculate the overall enthalpy change:
ΔH = -1244.4 kJ - 375.6 kJ - 517.0 kJ
ΔH = -2137 kJ
Therefore, the enthalpy change for the reaction N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l) is -2137 kJ.