Find the enthalpy for 10.0g of NH3 when it is made from molecular hydrogen and molecular nitrogen. The answer is -27.2kj but I'm confused about how to find enthalpy if its in grams.
N2 + 3H2 ==> 2NH3
Look up delta Ho in your text for N2 (it will be zero) and H2 (it will be zero) and NH3 (I don't remember that number). Then plug that numbers into
dHrxn = (n*dHo products) - (n*dHo reactants) and solve for dHrxn/ That dHrxn will be 1 mol two mols of NH3 (2*17 = 34 g). Convert to 10.0 grams.
Post your work if you get stuck.
To calculate the enthalpy for a reaction, you need to use the concept of stoichiometry and the enthalpies of formation of the reactants and products.
In this case, you are given the mass of NH3, which is 10.0g, and you want to find the enthalpy change for the reaction where NH3 is formed from molecular hydrogen (H2) and molecular nitrogen (N2).
To find the enthalpy change, you can follow these steps:
1. Write the balanced chemical equation for the reaction:
N2 (g) + 3H2 (g) → 2NH3 (g)
2. Determine the molar mass of NH3 by adding the atomic masses of nitrogen (N) and hydrogen (H):
Molar mass of NH3 = (1 mol of N × atomic mass of N) + (3 mol of H × atomic mass of H)
3. Convert the given mass of NH3 to moles by using its molar mass:
Moles of NH3 = mass of NH3 / molar mass of NH3
4. Use the balanced equation to find the moles of NH3 produced from the moles of N2:
Moles of NH3 = 2 × moles of N2
5. Calculate the enthalpy change (∆H) using the equation:
∆H = (moles of NH3 × ∆Hf(NH3)) - (moles of N2 × ∆Hf(N2))
The ∆Hf values are the enthalpies of formation and can be used to calculate the enthalpy change for a reaction. These values can be found in a reference table or textbook.
6. Substitute the values into the equation to calculate the enthalpy change:
∆H = (moles of NH3 × ∆Hf(NH3)) - (moles of N2 × ∆Hf(N2))
(∆Hf(NH3) is the enthalpy of formation of NH3
∆Hf(N2) is the enthalpy of formation of N2)
By following these steps and substituting the given values, you can calculate the enthalpy change (∆H) for the reaction.
Note: The negative sign (-27.2 kJ) indicates an exothermic reaction, which means heat is released during the reaction.