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RideAnS [48]
3 years ago
7

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

oundings when 2.25 moles of CO(g) react at standard conditions. S°surroundings = J/K
Chemistry
1 answer:
Marizza181 [45]3 years ago
3 0

<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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Lactose, C12H22O11, is a naturally occurring sugar found in mammalian milk. A 0.335 M solution of lactose in water has a density
Dmitry_Shevchenko [17]

The molality of the solution is 0.00037 m.

<h3>What is concentration?</h3>

The term concentration refers to the amount of solute in a solution.

We have the following information;

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The molality of the solution is obtained from;

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Compare the root mean square speeds of O2 and UF6 at 65 degree Celsius
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<h3>Answer:</h3>

The root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.

<h3>Solution and Explanation:</h3>
  • To find how fast molecules or particles of gases move at a particular temperature, the root  mean square speed is calculated.
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       Root mean square=\sqrt{ \frac{3RT}{M}

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Molar gas constant = 8.3145 J/k.mol

Root mean square speed = \sqrt\frac{(3)(8.3145)(338K)}{0.032}

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The molar mass of UF₆ is 352 g/mol or 0.352 kg/mol        

Root mean square speed = \sqrt\frac{(3)(8.3145)(338K)}{0.352}

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