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3 years ago

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In the film, Lavoisier said that the mass gained by the pipe plus the mass of the collected gas "exactly" equaled the lost mass

of water. If we believe Lavoisier to have behaved scientifically, how must you interpret that statement?

1 answer:

Nuetrik [128]3 years ago

8 0

Answer:

I would interpret the statement by using a formula.

Explanation:

In order to be scientifical in a research, the scientific method must be used, this means that in this case, Lavoisier should´ve follow a Hypothesis, Objectives, a Methodlogy, Results and made a Discussion and Conclusion.

To prove that the mass gained by the pipe plus the mass of the collected gas "exactly" equaled the lost mass of water, Lavoisier should´ve used this statement as a formula where it proves with numbers that is correct.

For example:

mg= mass gained by the pipe

mc=mass of the collected gas

ml=lost mass of water

mg+mc=ml

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$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

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Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

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Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

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$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

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