Question

The entropy (S-°)of the following substances are :

CH4(g)186.2JK−1mol−1CH4(g)186.2JK−1mol−1

O2(g)205.0JK−1mol−1O2(g)205.0JK−1mol−1

CO2(g)213.6JK−1mol−1CO2(g)213.6JK−1mol−1

H2O(l)69.9JK−1mol−1H2O(l)69.9JK−1mol−1

The entropy change (△S0)(△S0) for the reaction

CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH4(g)+2O2(g)→CO2(g)+2H2O(l)is :

(a)−312.5JK−1mol−1(b)−242.8JK−1mol−1(c)−108.1JK−1mol−1(d)−37.6JK−1mol−1

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Answer 1

Answer:c) -242.8

Explanation:delta S =change in entropy (entropy of reactant - product)

=(213.6+2×69.9)-(186.2+2×205)

353.4-596.2

= -242.8

Answer 2

Answer:

-242.8

Explanation:

Given:

CH4(g)186.2JK−1mol−1CH4(g)186.2JK−1mol−1

O2(g)205.0JK−1mol−1O2(g)205.0JK−1mol−1

CO2(g)213.6JK−1mol−1CO2(g)213.6JK−1mol−1

H2O(l)69.9JK−1mol−1H2O(l)69.9JK−1mol−1

Change in entropy implies how much randomness there is in the provided .system. Therefore,

Change in entropy(Δ S) =(entropy of reactant - entropy of product).

=(213.6+2×69.9)-(186.2+2×205)

353.4-596.2

= -242.8JKmol-1

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