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Leno4ka [110]
3 years ago
13

An unknown compound does not dissolve in water, is a nonelectrolyte and has a low melting point. Would it be an ionic compound?

Explain.
Chemistry
1 answer:
Leviafan [203]3 years ago
8 0
No, it's not an ionic compound.
The distinguishing properties of ionic compounds are that they dissolve in water (water's polarity pulls the ions apart), are electrolytes (dissociate into ions, which conduct electricity), and have high melting points (it takes a lot of energy to separate the ions without water or other polar compounds). Since the unknown compound has the opposite of all these properties, it's not ionic.
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Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
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<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

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)}]

For the given chemical reaction:

2CO(g)+O_2(g)\rightarrow 2CO_2(g)

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

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]

We are given:

\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = \frac{173.008}{1}\times 2.25=432.52J/K

Hence, the value of \Delta S^o for the surrounding when given amount of CO is reacted is 432.52 J/K

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