N2 (g) + 3H2 (g) → 2NH3(g) ΔH = -92.2 kJ
What mass of ammonia is theoretically produced if the above reaction released 168 kJ of heat?
I don't buy 31.02 g. The correct answer is approx 62 g NH3 or
92.2 kJ will produce 34 g NH3 so 168 will produce 34 x 168/92.2 = ?
OR, in your work you show 1.82 and that 1.82 x 2*17.03 is correct because there is a coefficient of 2 for NH3 in the reaction.
q = -168 kj
delta H= -92.2 kj/mol
n=?
n= -q/ delta H
n= -168 kj / -92.2
n= 1.82
n= m/Molar mass of NH3
m= n* MM
m= 1.82 moles * 17.03 g/mole
m= 31.02 g
Therefore, approx . 31.02 grams of NH3 is produced
To solve this problem, we will use the concept of stoichiometry and the given enthalpy change (∆H) of the reaction.
Step 1: Find the number of moles of ammonia produced.
According to the balanced chemical equation, it states that 1 mole of N2 reacts to produce 2 moles of NH3. Therefore, the moles of NH3 can be calculated using the following ratio:
1 mole N2 : 2 moles NH3
Step 2: Calculate the moles of NH3 using the stoichiometry.
We can use the following equation:
moles NH3 = (moles of N2 * ∆H of reaction) / ∆H of NH3
Let's calculate this now.
Given: ΔH = -92.2 kJ
ΔH' (heat released) = 168 kJ
From the equation, we know that ΔH = -92.2 kJ corresponds to the production of 2 moles of NH3.
So, moles NH3 = (moles N2 * -92.2 kJ) / -92.2 kJ
Step 3: Calculate the moles of N2 using ΔH'.
Using the same equation, we can now solve for moles N2:
moles N2 = (moles NH3 * ∆H of NH3) / ∆H of reaction
moles N2 = (moles NH3 * -92.2 kJ) / 168 kJ
Step 4: Convert moles of NH3 to grams using molar mass.
The molar mass of NH3 is 17.03 g/mol.
mass NH3 = moles NH3 * molar mass of NH3
Finally, let's plug the numbers into the equations and calculate the mass of NH3.
mass NH3 = (moles NH3 * molar mass of NH3)
= ( (moles N2 * -92.2 kJ) / -92.2 kJ ) * molar mass of NH3
Using the balanced equation, we have determined that 1 mole of NH3 corresponds to the production of 2 moles of N2.
Now, let's calculate the values using the given information.
To find the mass of ammonia produced, we need to use the balanced equation and the enthalpy change value to calculate the amount of substance involved in the reaction.
Firstly, we need to determine the balanced chemical equation for the reaction:
N2 (g) + 3H2 (g) → 2NH3(g)
From the balanced equation, we can see that for every 1 mole of N2, we can produce 2 moles of NH3. Now, we need to use the enthalpy change value to calculate the amount of substance involved in the reaction.
Given: ΔH = -92.2 kJ
We have the energy change (-92.2 kJ) for the formation of 2 moles of NH3.
To find the energy change for the formation of 1 mole of NH3, we can divide the energy change by the stoichiometric coefficient:
Energy change for 1 mole of NH3 = (-92.2 kJ) / 2 = -46.1 kJ
Therefore, the energy change for the formation of 1 mole of NH3 is -46.1 kJ.
Now, we can use the energy change to find the moles of NH3 produced when 168 kJ of heat is released:
Moles of NH3 = Energy change / Energy change for 1 mole of NH3
Moles of NH3 = 168 kJ / -46.1 kJ
Moles of NH3 ≈ -3.64 moles
Since the reaction is proceeding in the forward direction, we can ignore the negative sign. Therefore, the number of moles of NH3 produced is approximately 3.64 moles.
Finally, we can calculate the mass of NH3 produced by using its molar mass:
Molar mass of NH3 = 14.01 g/mol (N) + 1.01 g/mol (H) × 3
Molar mass of NH3 = 17.03 g/mol
Mass of NH3 = Moles of NH3 × Molar mass of NH3
Mass of NH3 = 3.64 moles × 17.03 g/mol
Mass of NH3 ≈ 61.9 g
Therefore, the theoretical mass of ammonia produced is approximately 61.9 grams.