Determining Delta S of Cl2 + Br2 --> 2BrCl

[qwexor]

Hello Everyone,

I'm reviewing for my chem 12 midterm and just came across a question that I'm not sure about. If I have the following Reaction:


Cl2(g) + Br2(g) <---> 2BrCl (g)

How can i determine whether the change in entropy (Delta S), is positive or negative.

Thanks!,
Nick

 

[mrjeffy321]

you can look up the entropy values for each substance and then find the difference between the two sides to find the change in entropy.

 

[The Bob]

Cl2(g) + Br2(g) <---> 2BrCl (g)

How can i determine whether the change in entropy (Delta S), is positive or negative.

Well which side will have the higher entropy? That is the first question.

I think:The left side will have a higher entropy than the right side because there are more possible arrangments for the molecules on the left hand side than there are on the right hand side (as this is just one compound).

I, therefore, think it is negative because the entropy has decreased.[/color] (in white so that qwexor has the option of looking or not).

The Bob (2004 ©)

 

[gravenewworld]

Your are thinking way too hard. Do like bob said. How many different molecules do you have on the left side of the equation? How many different molecules are on the right side of the equation. 2 on the left and 1 on the right, therefore entropy has decreased. See how easy that was? Don't let your education interfere with your intelligence.

 

[Gokul43201]

The above solutions work best if you exclude \Delta S _{mixing} which may be of the same order, though opposite sign as \Delta S _{reaction}.

 

[GCT]

There seems to be two molecules for two molecules, as the right has the coefficient of 2. The molar ratio of products to ratio is the same. For this problem you might need to browse through the standard entropy values. It might be the case that the reactant side, entropy is zero. I'll have to read up more.