Calculate Delta(H) for the following: NH 3 (g) + HCl(g) --> NH 4 Cl(s) Given: 1/

Question

Calculate Delta(H) for the following:

NH3 (g) + HCl(g) --> NH4Cl(s)

Given:

1/2 N2(g) + 1.5H2(g) --> NH3(g) Delta(H)final= -45.8 kJ

1/2 H2(g) + 1/2Cl2(g) --> HCl(g) Delta(H)final= -92.3 kJ

NH3(g) --> NH3(aq) Delta(H)= -35.4 kJ

HCl(g) --> HCl(aq) Delta(H)= -75.1 kJ

We conducted an experiment in 2 phases, Neutralization reaction between HCl and NH3 and graphed the Delta(T) for a course of 10 minutes; we ran 2 trails. 2nd phase included an enthalpy of formation of NH4Cl in DI water; we ran 2 trails. The measurements and information we collected were as follows:

Experiment 1)

50mL of HCl; introduced 50mL of NH3 in both experiments.

Experiment 2)

100mL of DI water, introduced 5.491g of NH4Cl in first trial, 5.487g of NH4Cl in second trial.

I am completely lost; where do I start? Do I have everything I need to claculate Delat(H) for each experiment?

Explanation / Answer

Calculate Delta(H) for the following:

NH3 (g) + HCl(g) --> NH4Cl(s)

We know

Also

NH3(aq) + HCl(aq) -> NH4Cl(aq) -52.8kJ ...............1

NH4Cl(s) --> NH4Cl(aq) + 8.2kJ .........................2

Given:

1/2 N2(g) + 1.5H2(g) --> NH3(g) Delta(H)final= -45.8 kJ .......3

1/2 H2(g) + 1/2Cl2(g) --> HCl(g) Delta(H)final= -92.3 kJ .......4

NH3(g) --> NH3(aq) Delta(H)= -35.4 kJ ...........5

HCl(g) --> HCl(aq) Delta(H)= -75.1 kJ .........6

Answer

We can get the desired equation by following relation

So Hf for NH4Cl = eq'n 1 + (-eq'n 2) + eq'n 5 + eq'n 6 = -171.5kJ

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