Asked by Lauren
Can you please check my work? I got -4.3KJ/mol
If this is not right can you tell me what I am doing wrong? Thank you!
Questions. 1.a. Use Hess's Law and the measured mean enthalpy changes for the NaOH-HCl and NH3-HCl reactions to calculate the enthalpy change to be expected for the reaction:
NaOH + NH4Cl => NaCl + NH3 + H2O
It's wanting the delta H value.
These are the values from the mean enthalpy.
Delta H for NaOH & HCl -59 KJ/mol
Delta H for NH3 & HCl -50. KJ/mol
Delta H for NaOH & NH4Cl -5.7 KJ/mol
Answers
Answered by
DrBob222
I'm confused by your data.
You've listed dH for the reaction you want as -5.7 kJ/mol. So
a. why do you have the question if the answer is -5.7?
b. Why do you have -4.3 if the data says it is -5.7
If I did this and used ONLY the first two equations you have, I get this.
NaOH + HCl ==> NaCl + H2O dH = -59 kJ/mol
Reverse equation 2 to obtain
NH4Cl ==> HCl + NH3 dH = +50
Add the two equations
NaOH + NH4Cl ==> NaCl + H2O + NH3 and add dH -59 kJ/mol + 50 kJ/mol = -9 kJ/mol
You've listed dH for the reaction you want as -5.7 kJ/mol. So
a. why do you have the question if the answer is -5.7?
b. Why do you have -4.3 if the data says it is -5.7
If I did this and used ONLY the first two equations you have, I get this.
NaOH + HCl ==> NaCl + H2O dH = -59 kJ/mol
Reverse equation 2 to obtain
NH4Cl ==> HCl + NH3 dH = +50
Add the two equations
NaOH + NH4Cl ==> NaCl + H2O + NH3 and add dH -59 kJ/mol + 50 kJ/mol = -9 kJ/mol
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