Is \(\Delta S\) for the following process positive or negative? Why? \(\mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

Short Answer

Expert verified
The \(\Delta S\) for the given process is positive because the number of gaseous molecules increases from reactants to products.

Step by step solution

01

Understand the reaction

First, we analyze the given reaction. Here, one molecule of \(\mathrm{C}_{4}\mathrm{H}_{8}\) gas is decomposed into two molecules \(\mathrm{C}_{2}\mathrm{H}_{4}\) gas.
02

Determine the change in gaseous molecules

Next, we find the net change in gaseous molecules. On the left side (reactants) of the reaction there is 1 gaseous molecule while on the right side (products) there are 2 gaseous molecules. So the net change is 2 - 1 = 1.
03

Determine the sign of \(\Delta S\)

Entropy, in simple terms can be thought of as the measure of chaos or disorder in a system. An increase in the number of gaseous molecules results in an increase in disorder. Hence, since the number of gaseous molecules increases from reactants to products, the entropy of the system, \(\Delta S\), is positive.

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