With a platinum catalyst, ammonia will burn in oxygen to give nitric oxide, NO.

4NH3 + 5O2 -----> 4NO + 6H2O
Delta H= -906 kJ

What is the enthalpy change for the following reaction?

NO + 3/2H2O ---> NH3 + 5/4O2

Note that if you divide all the cofficients by 4 you will have

NH3 + 5/4 O2 ==> NO + 3/2 H2O
so delta H for that reaction is what? &nbsp&nbsp -906 kJ/4 ??
Then the reverse direction is just the negative of that.

To determine the enthalpy change for the given reaction, we can use the enthalpy change of the balanced reaction provided and apply Hess's Law.

The given reaction is:

4NH3 + 5O2 -----> 4NO + 6H2O (ΔH = -906 kJ)

We need to calculate the enthalpy change for the reaction:

NO + 3/2H2O ---> NH3 + 5/4O2

To find the enthalpy change for this reaction, we can reverse the reaction and change the sign of the enthalpy change:

NH3 + 5/4O2 ---> NO + 3/2H2O (ΔH = +906 kJ)

Next, we can multiply the reaction by a factor to match the coefficients:

2(NH3 + 5/4O2) ---> 2(NO + 3/2H2O) (ΔH = +906 kJ)

Now, we can compare the modified reaction to the target reaction:

4(NH3 + 5/4O2) = 4NO + 6H2O

We can see that the modified reaction is equivalent to two copies of the target reaction. Therefore, we can divide the enthalpy change by 2:

(ΔH for target reaction) = (+906 kJ) / 2 = +453 kJ

So, the enthalpy change for the reaction:

NO + 3/2H2O ---> NH3 + 5/4O2

is +453 kJ.

To determine the enthalpy change for the reaction, we can use the concept of Hess's law. According to Hess's law, the total enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps that make up the reaction.

In this case, we can use the given reaction and the reverse of the given reaction to break it down into two steps:
1. Reverse the given reaction:
NH3 + 5/4O2 ---> NO + 3/2H2O

2. Combine the two reactions to cancel out any common species:
4NH3 + 5O2 ---> 4NO + 6H2O
NH3 + 5/4O2 ---> NO + 3/2H2O

Next, add these two reactions together by aligning them and adding the corresponding coefficients:
4NH3 + 5O2 + NH3 + 5/4O2 ---> 4NO + 6H2O + NO + 3/2H2O

Simplifying this equation, we get:
5NH3 + 21/4O2 ---> 5NO + 15/2H2O

Now that we have a balanced equation, the coefficients represent the stoichiometric ratios of the reactants and products. The coefficient of NH3 is 5 in the final equation, while in the given reaction it is 1. This means we need to multiply the enthalpy change of the given reaction by 5 to match the coefficients:
Delta H = 5 * (-906 kJ) = -4530 kJ

Therefore, the enthalpy change for the reaction NO + 3/2H2O ---> NH3 + 5/4O2 is -4530 kJ.