which has more entropy
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which has more entropy
If there was a reaction where there were two moles of gas converted to one mole of gas but the product is a larger molecule than either of the reactants, would there be an increase in entropy or decrease?
For example:
A2(g) + B2(g) --> A2B2(g)
For example:
A2(g) + B2(g) --> A2B2(g)
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Re: which has more entropy
All gases act the same so even though the product is larger than either reactant, it doesn't matter. For example, CH3(g) acts the same as H2(g) even though it larger. In this case it would be an entropy decrease because moles of gas is decreasing.
Re: which has more entropy
The entropy would decrease because the moles of the reaction decrease. This occurs even if the molecule is larger.
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Re: which has more entropy
No matter whether a molecule is larger or smaller every mole of gas no matter the element or substance is going to have 6.02 x 10^23 particles. This means the entropy would always be decreasing since 1 mole of gas will have less particles than the two moles of gas.
Re: which has more entropy
Seeing that the moles decrease, this tells us that the entropy will decrease as well.
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Re: which has more entropy
the entropy would decrease in this case because the moles of the gas in the reaction decrease
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Re: which has more entropy
The reactants would have a higher entropy as there are more moles of gas. Even if the molecule is larger it still has fewer moles of gas meaning that the entropy will be smaller.
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Re: which has more entropy
The number of moles of gas decrease from 2 moles to 1 mole, so the entropy decreases.
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Re: which has more entropy
entropy in a reaction like this is dependent on the moles. So if moles decrease from the R to P then the entropy will also decrease
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Re: which has more entropy
Hello! As others said, the entropy would decrease because the number of moles of gas decrease from reactants to products. To add on, this is because with fewer moles, there can be more disorder or differing orientations of the molecules, and thus the entropy would increase. I hope this was helpful!
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