Calculate the enthalpy for the following reaction: 4 NH3 (g) + 5 O2 (g) into 4 NO (g) + 6 H2O (g).

You may only use the following information:
N2 (g) + O2 (g) into 2 NO (g); Delta Hf = 180.6 kJ
N2 (g) + 3 H2 (g) into 2 NH3 (g); Delta Hf = -91.8 kJ
2 H2 (g) + O2 (g) into 2 H2O (g); Delta Hf = -483.7 kJ

My answer was -910.6 KJ but the real answer was -918.3 KJ. I need to know what I am doing wrong.

I didn't get either answer.

Look at your equation to make sure it adds to the desired equation in the problem. Here is what I did.
equation 1 x 2
equation 2 reversed and x 2.
equation 3 x 3.
Add it. First look at the equation.
(2N2) + 2O2 + 4NH3 + (6H2) +3O2 ==>
(2N2) + + (6H2) + 4NO + 6H2O

The 2N2 and 6H2 cancel since they are on opposite sides of the equation and we are left with
4NH3 + 5O2 ==>4NO + 6H2O which is exactly what we want.
Then (180.6*2)+(91.8*2)+(-483.7*3) = -906.3 kJ.
Check my work.

Determine the enthalpy of formation of the reaction: [4 NH₃ + 5 O₂ -> 4 NO + 6 H₂O]

Well, it seems like you got close to the right answer, but missed a small detail. Let me break it down for you and show you where you went wrong.

First, let's write down the given equations:
1. N2 (g) + O2 (g) -> 2 NO (g); ΔHf = 180.6 kJ
2. N2 (g) + 3 H2 (g) -> 2 NH3 (g); ΔHf = -91.8 kJ
3. 2 H2 (g) + O2 (g) -> 2 H2O (g); ΔHf = -483.7 kJ

Now, let's manipulate the given equations to match the desired reaction:
1. 2 NO (g) -> N2 (g) + O2 (g) ; ΔHf = -180.6 kJ (flipped equation 1 and changed the sign)
2. 2 NH3 (g) -> N2 (g) + 3 H2 (g) ; ΔHf = 91.8 kJ (flipped equation 2 and changed the sign)
3. 2 H2O (g) -> 2 H2 (g) + O2 (g) ; ΔHf = 483.7 kJ (flipped equation 3 and changed the sign)

Now, let's add up the manipulated equations:
4. 4 NO (g) + 6 H2O (g) -> 2 N2 (g) + 3 O2 (g) ; ΔHf = -724.8 kJ (multiplied equation 1 by 2 and equation 3 by 3, then added them together)

Since we want the desired reaction to be:
4 NH3 (g) + 5 O2 (g) -> 4 NO (g) + 6 H2O (g),

we can simply add equation 2 to equation 4:
4 NO (g) + 6 H2O (g) + 4 NH3 (g) -> 2 N2 (g) + 3 O2 (g) + 4 NO (g) + 6 H2O (g) ; ΔHf = -633.0 kJ

Now, we can see that the coefficients of NO and H2O are equal on both sides of the equation, so we can cancel them out:
4 NH3 (g) + 5 O2 (g) -> 2 N2 (g) + 3 O2 (g) ; ΔH = -633.0 kJ

But wait, we want the ΔH for the given reaction, not per mole. So we have to divide the enthalpy change by the number of moles involved in the manipulated equation:
-633.0 kJ / (4/2) = -316.5 kJ

Well, it seems like you didn't account for the stoichiometric coefficients of the manipulated equation. Hence, you obtained a wrong value for the enthalpy change. The correct answer should be -316.5 kJ, not -910.6 kJ.

I hope this clears things up for you, and don't worry, chemistry can be tricky sometimes!

To calculate the enthalpy change for the given reaction, we can use the Hess's Law, which states that the overall enthalpy change in a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.

Given:
N2 (g) + O2 (g) -> 2 NO (g); ΔHf = 180.6 kJ
N2 (g) + 3 H2 (g) -> 2 NH3 (g); ΔHf = -91.8 kJ
2 H2 (g) + O2 (g) -> 2 H2O (g); ΔHf = -483.7 kJ

We need to manipulate and combine these equations to get the desired reaction:
4 NH3 (g) + 5 O2 (g) -> 4 NO (g) + 6 H2O (g)

Let's break down the desired reaction into multiple steps using these given equations:

1. 4 NH3 (g) -> 4/2 N2 (g) + 4/2 H2 (g)
Since we need 4/2 moles of N2 and H2, we multiply the second equation by 2 and reverse it to get:
2 NH3 (g) -> N2 (g) + 3 H2 (g)
ΔH1 = 2 * ΔHf = 2 * (-91.8 kJ) = -183.6 kJ (Multiplied by 2 because we need 2 moles of NH3)

2. 5 O2 (g) -> 10/2 O2 (g)
Since we need 10/2 moles of O2, we multiply the first equation by 5:
5 N2 (g) + 5 O2 (g) -> 10/2 NO (g)
ΔH2 = 5 * ΔHf = 5 * (180.6 kJ) = 903 kJ (Multiplied by 5 because we need 5 moles of O2)

3. 10/2 NO (g) + 10 H2 (g) -> 10/2 NO (g) + 10/2 H2O (g)
Since we need 10/2 moles of NO and H2O, we multiply the third equation by 10/2:
10 NO (g) + 10 H2 (g) -> 10/2 H2O (g)
ΔH3 = (10/2) * ΔHf = (10/2) * (-483.7 kJ) = -2418.5 kJ (Multiplied by 10/2 because we need 10 moles of NO and H2O)

Now, we can sum up these individual enthalpy changes to find the enthalpy change of the desired reaction:

ΔH_total = ΔH1 + ΔH2 + ΔH3
= -183.6 kJ + 903 kJ + (-2418.5 kJ)
= -1699.1 kJ

So, the enthalpy change for the given reaction is -1699.1 kJ, which is different from the value you obtained (-910.6 kJ).

Please recheck your calculations to identify the source of the error.

To calculate the enthalpy for the given reaction, we can use the concept of Hess's Law, which states that the total enthalpy change of a reaction is equal to the sum of enthalpy changes of individual steps within the reaction.

In this case, we want to obtain the enthalpy change for the reaction: 4 NH3 (g) + 5 O2 (g) -> 4 NO (g) + 6 H2O (g)

First, we need to manipulate the given equations to match the desired reaction equation. Let's reverse the second equation to match the NH3 and NO compounds:

2 NH3 (g) -> N2 (g) + 3 H2 (g) (Reversed: -(-91.8 kJ) = +91.8 kJ)

Next, we need to multiply the first equation by 4 to balance the NO and NH3 coefficients:

4(N2 + O2 -> 2 NO) (4 * 180.6 kJ = 722.4 kJ)

Now, let's multiply the third equation by 3 to match the H2O coefficient:

6(H2 + 1/2 O2 -> H2O) (3 * (-483.7 kJ) = -1451.1 kJ)

Now, we sum up these equations to get our desired reaction equation:

4 NH3 (g) + 5 O2 (g) -> 4 NO (g) + 6 H2O (g)
= 2 NH3 - N2 - 3 H2 + 4(N2 + O2 -> 2 NO) + 3(H2 + 1/2 O2 -> H2O)
= 2 NH3 - N2 - 3 H2 + 4NO + 5/2 O2 + 6H2O - 3/2O2
= 2 NH3 - N2 + 4NO + 6H2O - 3 H2

Now, we sum the enthalpy changes of the individual steps:

Enthalpy change for the desired reaction = (+91.8 kJ) + (722.4 kJ) + (-1451.1 kJ)

Calculating this sum, we obtain:

Enthalpy change = -636.9 kJ

Therefore, the correct answer for the enthalpy change of the given reaction is -636.9 kJ, not -918.3 kJ.

It appears there was an error in the processing of the given information. Double-check your calculations and ensure you have accurately inputted the values and signs for each given reaction.