yazyj
D89B851E-B633-4996-8085-BCDD197BE59E.jpeg

Why is option 1) wrong? What does “nucleons” refer to here?  
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yazyj

Did the entropy of 1) increase because solid state becomes gaseous state? And also what are the factors that affect entropy other than state and no of gaseous molecules? 

For the number of gaseous molecules do we look at the number of atoms or number of moles (based off of the eqn)?

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BedokFunlandJC
yazyj wrote:
D89B851E-B633-4996-8085-BCDD197BE59E.jpeg

Why is option 1) wrong? What does “nucleons” refer to here?  


Nucleons = particles in the neutrons = protons + neutrons.

Since 58 < 60, so less nucleons, not more.
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BedokFunlandJC
yazyj wrote:

Did the entropy of 1) increase because solid state becomes gaseous state? And also what are the factors that affect entropy other than state and no of gaseous molecules? 

For the number of gaseous molecules do we look at the number of atoms or number of moles (based off of the eqn)?

79FEA51E-E732-46CE-9339-47A59CFA4C76.jpeg 


Option 1 : Yes, because solid S is being combusted to generate gaseous SO2.

For H2 Chem purposes, change of state is the most important, followed by change in amount of gaseous particles from LHS to RHS. That's it. No other factors required to be considered at A levels, ceteris paribus (eg. equal temperature, volume, pressure, etc). That is to say, if the equation is exactly the same for 2 options, the option with the higher temperature, larger volume or lower pressure, has greater entropy.

Look at amount (in moles) of particles (eg. atoms, molecules, ions, etc) when comparing 2 different options, not necessarily in atoms per se.
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