Reaction: 1/2N2 + 3/2H2 --> NH3

Delta H = -46Kj/mol
1) Find delta H n(NH3)formed=5 mol. M(NH3=8.5g)
2) Find delta H
a)3 mol H2 + 3 mol N2
b) 3 mol H2 + 0.5 mol N2
M NH3 = 17 g/mol
???

I don't understand the question with your use of symbols. M stands for molarity but I think you mean molar mass NH3 is 17. I have no idea what M(NH3)= 8.5 g means nor H n(NH3) formed = 5 mol.

The mass of NH3 is 8.5 grams

And the number of moles of NH3 that was formed was 5 mol (n)
Any help? :)

DrBob222

To find the enthalpy change (ΔH) for each scenario, we can use the given enthalpy change for the reaction 1/2N2 + 3/2H2 -> NH3, which is -46 KJ/mol.

1) Find ΔH when 5 moles of NH3 are formed. The molar mass (M) of NH3 is given as 17 g/mol.
To calculate ΔH, we can use the following formula:
ΔH = (number of moles) * (ΔH per mole)

Given:
Number of moles of NH3 formed (n(NH3)) = 5 mol
Molar mass of NH3 (M(NH3)) = 17 g/mol

First, calculate the mass of NH3 formed using the given molar mass:
Mass of NH3 formed = n(NH3) * M(NH3)
Mass of NH3 formed = 5 mol * 17 g/mol
Mass of NH3 formed = 85 g

Now we can calculate the ΔH for this scenario:
ΔH = (mass of NH3 formed) / (molar mass of NH3) * (ΔH per mole)
ΔH = 85 g / 17 g/mol * -46 KJ/mol
ΔH = -230 KJ

Therefore, the ΔH when 5 moles of NH3 are formed is -230 KJ.

2) Find ΔH for the given scenarios:
a) 3 mol H2 + 3 mol N2
Since the reaction has coefficients of 1/2, we need to double the ΔH. Hence,
ΔH for a) = 2 * ΔH = 2 * -46 KJ/mol = -92 KJ

b) 3 mol H2 + 0.5 mol N2
To calculate the ΔH for this scenario, we need to account for the reaction coefficients. In the given reaction, 1/2N2 + 3/2H2 -> NH3, we can see that for every 3 mol of H2 required, we need 1 mol of N2. Therefore, we need to adjust the ΔH by a factor of 2.5 (3 / 1.2 = 2.5). Hence,
ΔH for b) = 2.5 * ΔH = 2.5 * -46 KJ/mol = -115 KJ

Therefore, the ΔH for the scenarios are as follows:
a) ΔH = -92 KJ
b) ΔH = -115 KJ