How do I find the molar enthalpy H (kJ/mol) for reaction of NH3 + HCl -> NH4Cl at temp 9.8 celsius by using Hess's Law. I do not understand why there are unknown variables and how to use that to find the molar enthalpy. PLEASE HELP.
1/2 N2 + 3/2 H2 + xH2o -> NH3 H= -80.7 kJ
1/2 H2 + 1/2 Cl2 + yH2o -> HCl H= -164.7 kJ
To find the molar enthalpy (H) for the reaction NH3 + HCl -> NH4Cl using Hess's Law, you can follow these steps:
Step 1: Write the balanced equation for the target reaction.
NH3 + HCl -> NH4Cl
Step 2: Identify the known reactions that have enthalpy changes (H) provided.
These are the two reactions given:
1/2 N2 + 3/2 H2 + xH2o -> NH3 H = -80.7 kJ (Reaction 1)
1/2 H2 + 1/2 Cl2 + yH2o -> HCl H = -164.7 kJ (Reaction 2)
Step 3: Manipulate the known reactions to obtain the target reaction.
To do this, you will multiply Reaction 1 by 2 and Reaction 2 by 2, so that the number of moles of NH3 and HCl in both reactions match the target reaction.
2 * [1/2 N2 + 3/2 H2 + xH2o -> NH3] H = 2 * (-80.7 kJ)
H2 + Cl2 + 2yH2o -> 2HCl H = 2 * (-164.7 kJ)
After manipulation, the equations become:
N2 + 3H2 + 2xH2o -> 2NH3 H = -161.4 kJ (Modified Reaction 1)
2H2 + 2Cl2 + 4yH2o -> 4HCl H = -329.4 kJ (Modified Reaction 2)
Step 4: Add the manipulated reactions together to obtain the target reaction.
N2 + 3H2 + 2xH2o + 2H2 + 2Cl2 + 4yH2o -> 2NH3 + 4HCl
By canceling out the common components:
N2 + 5H2 + 2xH2o + 2Cl2 + 4yH2o -> 2NH3 + 4HCl
Step 5: Determine the enthalpy change (H) for the target reaction.
Since enthalpy is a state function, the overall change in enthalpy is the sum of the individual changes.
Therefore, the enthalpy change for the target reaction is:
H(target) = H(Modified Reaction 1) + H(Modified Reaction 2)
= -161.4 kJ + (-329.4 kJ)
= -490.8 kJ
So, the molar enthalpy change (H) for the reaction NH3 + HCl -> NH4Cl at temperature 9.8 degrees Celsius, using Hess's Law, is -490.8 kJ/mol.