Given that 4NH3+5O2-4NO+6H2O ΔH= -906kJ.

What is the Enthalpy change for NO +3/2H2O - NH3 +5/2 O2

I don't know. Your second equation makes no sense. I think you must have meant 5/4 O2 and not 5/2. Assuming you meant 5/4, there are two steps to perform.

1. The equation is reversed; therefore, dH will tbe the negative of the forward reaction.
2. You've gone from 4 NO to NO and the new equation is 1/4 of of the old so take 1/4 of the new dH.

To find the enthalpy change for the reaction NO + 3/2 H2O → NH3 + 5/2 O2, we can use the given enthalpy change for the reverse reaction and then reverse the sign.

Given: 4 NH3 + 5 O2 → 4 NO + 6 H2O ΔH = -906 kJ

1. Apply Hess's Law: ΔH for reverse reaction = -ΔH for forward reaction.

So, ΔH for the reverse reaction: -(-906 kJ) = +906 kJ

2. Adjust the coefficients to match the desired reaction:

Multiply the given equation by 2 to cancel out the fractions:
8 NH3 + 10 O2 → 8 NO + 12 H2O ΔH = -1812 kJ

3. Flip the equation to get the desired reaction (NO + 3/2 H2O → NH3 + 5/2 O2):
8 NO + 12 H2O → 8 NH3 + 10 O2 ΔH = +1812 kJ

The enthalpy change for the reaction NO + 3/2 H2O → NH3 + 5/2 O2 is +1812 kJ.

To find the enthalpy change for the reaction NO + 3/2H2O - NH3 + 5/2O2 using the given reaction 4NH3 + 5O2 - 4NO + 6H2O, you need to apply the concept of Hess's Law.

Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken to reach the products, as long as the initial and final conditions are the same.

The first step is to rearrange the given reaction in order to match the target reaction:

4NH3 + 5O2 - 4NO + 6H2O

Multiply through by the stoichiometric coefficients to balance the equation:

8NO + 12H2O - 8NH3 + 20O2

To match the target reaction NO + 3/2H2O - NH3 + 5/2O2, multiply through by the appropriate coefficients:

4NO + 6H2O - 4NH3 + 10O2

Now, you can compare the coefficients between the target reaction and the derived equation to determine the enthalpy change. In this case, the derived equation has equal coefficients, so the enthalpy change remains the same:

ΔH = -906 kJ

Therefore, the enthalpy change for the reaction NO + 3/2H2O - NH3 + 5/2O2 is -906 kJ, which is the same as the given reaction.