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irga5000 [103]
3 years ago
7

Two samples of carbon tetrachloride were decomposed into their constituent elements. One sample produced 38.9 g of carbon and 45

1 g of chlorine, and the other sample produced 14.8 g of carbon and 135 g of chlorine. Are these results consistent with the law of definite proportions? Show why or why not.
Chemistry
1 answer:
Nataliya [291]3 years ago
3 0

Answer:

No.

Explanation:

The results mentioned are <u>not consistent </u>with the law of definite proportions.

Law of definite proportions states that a chemical compound contains exactly the same elements in the same proportion by weight independent of its source and method of it's preparation.

In the given question, the first sample contains 38.9 g of carbon and 451 g of chlorine that means the sample has 8% of carbon and 92% of chlorine.

On the other hand, second sample contains 14.8 g of carbon and 135 g of chlorine which means that this sample has 10% of carbon and 90% of chlorine.

Since the constituent elements in both the samples are not in fixed and constant proportions (by mass), the law of definite proportions fails here.

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<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of HCl gas is reacted is 73.21 J/K

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Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

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4HCl(g)+O_2(g)\rightarrow 2H_2O(g)+2Cl_2(g)

The equation for the entropy change of the above reaction is:  

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cl_2(g))})+(2\times \Delta S^o_{(H_2O(g))})]-[(4\times \Delta S^o_{(HCl(g))})+(1\times \Delta S^o_{(O_2(g))})]

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\Delta S^o_{(O_2(g))}=205.14J/K.mol\\\Delta So_{(HCl(g))}=186.91J/K.mol\\\Delta S^o_{(Cl_2(g))}=223.07J/K.mol\\\Delta S^o_{(H_2O(g))}=188.82J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (223.07))+(2\times (188.82))]-[(4\times (186.91))+(1\times (205.14))]\\\\\Delta S^o_{rxn}=-129J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-129) J/K = 129 J/K

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Moles of HCl gas reacted = 2.27 moles

By Stoichiometry of the reaction:

When 4 mole of HCl gas is reacted, the entropy change of the surrounding will be 129 J/K

So, when 2.27 moles of HCl gas is reacted, the entropy change of the surrounding will be = \frac{129}{4}\times 2.27=73.21J/K

Hence, the value of \Delta S^o for the surrounding when given amount of HCl gas is reacted is 73.21 J/K

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