Hess\' Law Enthalpy change for the decomposition of ammonium chloride 1. Write o

Question

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

Explanation / Answer

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.

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