Which substances in each of the following pairs would have the greater entropy? Explain.

(a) at 75°C and 1 atm: 1 mol H2O(l) or 1 mol H2O(g)

(b) at 5°C and 1 atm: 50.0 g Fe(s) or 0.80 mol Fe(s)

(c) 1 mol Br2(l,1 atm, 8°C) or 1 mol Br2(s, 1 atm, -8°C)

(d) 0.312 mol SO2 (g, 0.110 atm, 32.5°C) or 0.284 mol O2(g, 15.0 atm, 22.3°C)

My answers:

a) 1 mol of H2O(g) would have the higher entropy becasue since both items being compared are at the same temperature and pressure, we can determine which one has the higher entropy by choosing the one which has the state of greater microstates, that is gas between gas and solid states.

b) I'm not sure how to approach this part. If someone could help, it'd be appreciated.

c) 1 mol of Br2(l) at 1 atm and 8°C because at higher temperatures there are greater number of microstates and there are also greater number of microstates in a liquid state than a solid state.

d) not sure. please help!

b) if you convert 50g Fe to mol by divide by the molar mass you get 0.89 which is higher than the second choice

d) the higher temperature is the correct one

answering way too late but it will help for anyone who needs this later on

To determine which substances in each pair have greater entropy, we need to consider several factors:

(a) at 75°C and 1 atm: 1 mol H2O(l) or 1 mol H2O(g)
In this case, both substances are at the same temperature and pressure. Entropy is related to the number of microstates available to a system. In general, a substance in gas phase has higher entropy compared to the same substance in the liquid phase, as gas particles have more freedom to move and occupy a larger volume. Therefore, 1 mol of H2O(g) would have a greater entropy compared to 1 mol of H2O(l) at the given conditions.

(b) at 5°C and 1 atm: 50.0 g Fe(s) or 0.80 mol Fe(s)
To determine the entropy, it is useful to consider the molar mass of Fe, which is approximately 56 g/mol.
For 50.0 g of Fe(s), we can calculate the number of moles:

Number of moles (n) = Mass (m) / Molar Mass (M)
n = 50.0 g / 56 g/mol = 0.892 mol

So, we have 0.892 mol Fe(s) compared to 0.80 mol Fe(s). Since the molar quantity of Fe(s) is greater in the first case, it means that the first pair has a higher entropy.

(c) 1 mol Br2(l,1 atm, 8°C) or 1 mol Br2(s, 1 atm, -8°C)
Entropy generally increases with increasing temperature. So, at higher temperatures, the substance is likely to have a greater entropy compared to the same substance at lower temperatures. Additionally, in general, liquid state has higher entropy compared to the solid state due to more disorder within the liquid phase. Therefore, 1 mol of Br2(l) at 1 atm and 8°C would have higher entropy compared to 1 mol of Br2(s) at 1 atm and -8°C.

(d) 0.312 mol SO2 (g, 0.110 atm, 32.5°C) or 0.284 mol O2(g, 15.0 atm, 22.3°C)
To compare the entropies, we need to consider two main factors: molar quantity and pressure. Since both substances are gases, the quantity of moles is an important factor. However, pressure also plays a role, as higher pressure limits the available degrees of freedom for the gas particles, reducing its entropy.
To compare them properly, we need to calculate the standard entropy change (∆S°) for each substance and then compare them. Unfortunately, without the necessary data or equations, we cannot determine which substance has a higher entropy in this case.

b) To compare the entropy of 50.0 g Fe(s) and 0.80 mol Fe(s) at the same temperature and pressure, we need to consider the definition of entropy. Entropy is a measure of the number of microstates or arrangements of particles in a system. Since both samples are at the same temperature and pressure, the entropy can be compared based on the number of particles.

50.0 g Fe(s): To determine the number of moles in 50.0 g of Fe, we need to use the molar mass of Fe, which is approximately 55.85 g/mol.

Number of moles = mass (g) / molar mass = 50.0 g / 55.85 g/mol ≈ 0.894 mol

0.894 mol Fe(s) would have a higher entropy compared to 0.80 mol Fe(s) because there are more particles in 0.894 mol, indicating a greater number of microstates.

c) To compare the entropy of 1 mol Br2(l, 1 atm, 8°C) and 1 mol Br2(s, 1 atm, -8°C), we consider the states and temperatures of the substances.

At the same pressure, the entropy of a substance generally increases with temperature as there are more available microstates. So, the Br2(l) at 8°C would have a higher entropy compared to Br2(s) at -8°C.

d) To compare the entropy of 0.312 mol SO2(g, 0.110 atm, 32.5°C) and 0.284 mol O2(g, 15.0 atm, 22.3°C), we consider the number of particles and the states of the substances.

Since both substances are gases, we need to consider the number of particles (moles) and temperature. Comparing 0.312 mol SO2 and 0.284 mol O2, the higher number of moles indicates a greater number of microstates, suggesting that 0.312 mol SO2 would have a higher entropy. The pressure and temperature values provided are not relevant for comparing the entropy in this case.