Hess' Law
Enthalpy change for the decomposition of ammonium chloride
1. Write out the reaction NH4Cl(s) -> NH3 (g) + HCl(g) as a
series of steps which include the reactions observed in Procedures
2 and 3.
Use the known enthalpies for the change of state of NH3 and Hcl,
given below.
NH3 (g) -> NH3 (aq) (?H = -34,640 J/mol)
HCl (g) -> HCl (aq) (?H = -75,140 J/mol)
Be sure to show how the reaction steps must proceed so that delta H
for the desired reaction can be calculated. And be careful to use
the positive or negative enthalpy values depending on the direction
of the reactions that you add together.
I did this and got:
(A) NH3(g) -> NH3(aq) (-34640)
+(B) HCl(g) -> HCl(aq) (-75140)
+(D) HCl (aq) & NH3(aq) => NH4Cl(aq) (-3569.78)
gives: NH3 (g) & HCl (g) --> NH4Cl (aq)
-(C): NH4Cl (aq) --> NH4Cl (s) (2132.47)
(-34640)+(-75140)+(-3569.78) - (2132.47) = I
Answer �
We need to calculated the enthalpy change for the decomposition of ammonium chloride-
NH4Cl(s) -----> NH3 (g) + HCl(g), ?Hof = ?
We know the series of steps which include the reactions observed and standard enthalpies of each reaction as follow-
NH3(g) ----> NH3(aq) ?Hof = -34900 J/mol
HCl(g) ----> HCl(aq) ?Hof = - 74900 J/mol
HCl (aq) + NH3(aq) => NH4Cl(aq) ?Hof = -53000 J/mol
NH4Cl (aq) --> NH4Cl (s) ?Hof = - 14550 J/mol
When we added all these reaction we will get
NH3 (g) + HCl(g) ----> NH4Cl(s) ?Hof = -177350 J/mol
So when we reverse this reaction the sign of the ?Hof also get reversed and reaction is �
NH4Cl(s) ----> NH3 (g) + HCl(g) , ?Hof = + 177.350 kJ/mol
So correct answer for the enthalpy change for the decomposition of ammonium chloride is + 177.350 kJ/mole
So it is soo close with standard enthalpy value from test book. The standard enthalpy value for the decomposition of ammonium chloride is +176.34 kJ/mole.
Hess' Law Enthalpy change for the decomposition of ammonium chloride 1. Write out the reaction NH4Cl(s)...
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