Hess¡¯ law says that when we add reactions together we can also add the enthalpies. Given the following set of reactions, use Hess¡¯ Law to calculate the enthalpy of the reaction given below.

Question

Hess¡¯ law says that when we add reactions together we can also add the enthalpies. Given the following set of reactions, use Hess¡¯ Law to calculate the enthalpy of the reaction given below.

N2H4 + H2 ¡æ 2 NH3 ¥ÄH¡Æ = ?
N2 + 3 H2 ¡æ 2 NH3 ¥ÄH¡Æ = -92.38 kJ/mol
N2H4 + O2 ¡æ N2 + 2 H2O ¥ÄH¡Æ = -622.09 kJ/mol
2 H2 + O2 ¡æ 2 H2O ¥ÄH¡Æ = -483.6 kJ/mol

Answer 1

There are symbols which I don't understand in this problem that you typed, like these ¡æ and ¥ÄH¡Æ. But I understand the reactions. :3

We need to find the ΔH for this reaction:
N2H4 + H2 ---> 2 NH3

using the data for these reactions:
N2 + 3 H2 ---> 2 NH3 ; ΔH = -92.38 kJ/mol
N2H4 + O2 ---> N2 + 2 H2O ; ΔH = -622.09 kJ/mol
2 H2 + O2 ---> 2 H2O ; ΔH = -483.6 kJ/mol

What we'll do is reverse the third reaction. By doing this, we change the sign of its ΔH to its opposite. And then we add all the reactions, as well as the ΔH's.
N2 + 3 H2 ---> 2 NH3 ; ΔH = -92.38 kJ/mol
N2H4 + O2 ---> N2 + 2 H2O ; ΔH = -622.09 kJ/mol
2 H2O ---> 2 H2 + O2 ; ΔH = +483.6 kJ/mol
---------------------------------------------------------
N2H4 + H2 ---> 2 NH3 ; ΔH = 230.87 kJ/mol

hope this helps~ `u`

Answer 2

jae means ==>

i followed by upper case _ is '
That Ything is delta H = ?

Answer 3

To calculate the enthalpy of the reaction N2H4 + H2 -> 2 NH3, we can use Hess's Law by manipulating and combining the given reactions.

First, let's rearrange the equations and their enthalpy values so that the reactants on both sides match:

1. N2H4 + O2 -> N2 + 2 H2O ΔH° = -622.09 kJ/mol
2. N2 + 3 H2 -> 2 NH3 ΔH° = -92.38 kJ/mol
3. 2 H2 + O2 -> 2 H2O ΔH° = -483.6 kJ/mol

Now, we want to manipulate these equations in order to cancel out the desired reaction:

1. N2H4 + O2 -> N2 + 2 H2O ΔH° = -622.09 kJ/mol
2. 2 NH3 -> N2 + 3 H2 ΔH° = +92.38 kJ/mol

Note that we flipped the sign of the enthalpy value for reaction 2 because we are using it in the reverse direction.

The next step is to multiply the equations by a factor to balance the number of moles of the desired reaction:

1. 2 NH3 -> 2 N2 + 3 H2O ΔH° = +184.76 kJ/mol (2 times reaction 2)
2. N2H4 + O2 -> N2 + 2 H2O ΔH° = -622.09 kJ/mol (1 times reaction 1)

Now, we can add these two equations together to get the balanced equation and the enthalpy change for the reaction N2H4 + H2 -> 2 NH3:

N2H4 + O2 -> 2 N2 + 2 H2 + 3 H2O ΔH° = -437.33 kJ/mol

So, the enthalpy change for the reaction N2H4 + H2 -> 2 NH3 is -437.33 kJ/mol.