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
???
See your post above.
To find the enthalpy change (ΔH) when 5 moles of NH3 are formed and the molar mass of NH3 is 17 g/mol, you can use the following steps:
1) Convert the mass of NH3 (8.5 g) to moles:
Molar mass of NH3 = 17 g/mol
Moles of NH3 = Mass of NH3 / Molar mass of NH3
Moles of NH3 = 8.5 g / 17 g/mol
Moles of NH3 = 0.5 mol
2) Since the balanced equation shows that 1/2 N2 reacts with 3/2 H2 to form 1 mole of NH3, we can calculate the moles of N2 and H2 required using the mole ratios:
a) For 3 moles of H2 and 3 moles of N2:
Moles of NH3 = 5 mol (given)
Moles of H2 = 3 mol
Moles of N2 = 3 mol
b) For 3 moles of H2 and 0.5 moles of N2:
Moles of NH3 = 5 mol (given)
Moles of H2 = 3 mol
Moles of N2 = 0.5 mol
3) Calculate the ΔH for each scenario using the given ΔH value of -46 kJ/mol:
a) For 3 moles of H2 and 3 moles of N2:
ΔH = ΔH per mole of NH3 x Moles of NH3
ΔH = -46 kJ/mol x 5 mol
ΔH = -230 kJ
b) For 3 moles of H2 and 0.5 moles of N2:
ΔH = ΔH per mole of NH3 x Moles of NH3
ΔH = -46 kJ/mol x 5 mol
ΔH = -230 kJ
Both scenarios have the same enthalpy change, which is -230 kJ.
To find delta H for the given reactions, we need to use the given equation and the stoichiometry of the reactions.
1) Find delta H for the formation of 5 mol of NH3 (M(NH3) = 8.5 g).
Step 1: Calculate the moles of NH3 based on the given mass:
Moles of NH3 = mass / molar mass
= 8.5 g / 17 g/mol
= 0.5 mol
Step 2: Use the given delta H to calculate the delta H for the formation of NH3:
Delta H = Mols of NH3 x Delta H per mole of NH3 formed
= 0.5 mol x (-46 kJ/mol)
= -23 kJ
Therefore, the delta H for the formation of 5 mol of NH3 is -23 kJ.
2) Find delta H for the reactions of:
a) 3 mol H2 + 3 mol N2
b) 3 mol H2 + 0.5 mol N2
To find the delta H for a reaction, we can use Hess's Law.
a) To find delta H for 3 mol H2 + 3 mol N2:
Since the given reaction has the same stoichiometry as the balanced equation, the delta H remains the same as the given value: -46 kJ.
b) To find delta H for 3 mol H2 + 0.5 mol N2:
We need to use the concept of thermochemical equations and Hess's Law.
Step 1: Write the balanced equation for the given reaction:
1/2N2 + 3/2H2 --> NH3
(Note: the coefficients in the balanced equation are fractions.)
Step 2: Multiply the entire equation by 2 to get rid of the fractions:
N2 + 3H2 --> 2NH3
Step 3: Determine the delta H for the reaction based on the balanced equation:
Delta H = 2 x (-46 kJ)
= -92 kJ
Therefore, the delta H for the reaction 3 mol H2 + 0.5 mol N2 is -92 kJ.
To summarize:
a) The delta H for the reaction 3 mol H2 + 3 mol N2 is -46 kJ.
b) The delta H for the reaction 3 mol H2 + 0.5 mol N2 is -92 kJ.