[Solved] Can you please look over these questions

Answered step by step

Verified Expert Solution

Link Copied!

Question

1 Approved Answer

Posted on Sep 23, 2024

Can you please look over these questions and make sure I am correct if you do what is asked thumbs up if not then i

Can you please look over these questions and make sure I am correct if you do what is asked thumbs up if not then i will thumbs down:

Focus 3 - Discussion 3.4, 3.6; Exercises 3A.1 (just a), 3A.3, 3B.4, 3B.9, 3C.3, 3D.2 (a, b, c, 15 points)

3.4 Justify the identification of the statistical entropy with the thermodynamic entropy.

Although statistical entropy and thermodynamic entropy are two distinct ideas, they are connected and can be confused with one another in some situations.

A key idea in thermodynamics is the term "thermodynamic entropy," which describes how chaotic or random a system is. It is a macroscopic characteristic that can be measured through experimentation and is connected to the heat that is transferred during a reversible process between a system and its surroundings.

The number of microstates that are consistent with a specific macrostate of a system is related to the statistical entropy concept in statistical mechanics, on the other hand. It is a microscopic property that can theoretically be calculated but cannot be directly measured.

The Boltzmann entropy formula, which connects a system's statistical entropy S stat to its thermodynamic entropy S thermo, justifies the association between statistical and thermodynamic entropies:

S stat = ln Wk B

where W is the number of system microstates that are consistent with its macrostate and k B is the Boltzmann constant. This equation demonstrates that the statistical entropy, a measurement of the level of disorder or randomness in the system, is proportional to the logarithm of the number of microstates.

The statistical entropy approaches the thermodynamic entropy in the thermodynamic limit, where there are very many particles in the system, and the Boltzmann formula becomes:

S thermo equals k B ln

where n is the system's total number of accessible states. This equation demonstrates that the thermodynamic entropy, a measure of the degree of disorder or randomness in the system, is also proportional to the logarithm of the number of accessible states.

The Boltzmann formula offers a theoretical justification for this identification, so in the thermodynamic limit, the statistical entropy can be equated with the thermodynamic entropy.

3.6 The evolution of life requires the organization of a very large number of molecules into biological cells. Does the formation of living organisms violate the Second Law of thermodynamics? State your conclusion clearly and present detailed arguments to support it.

Because the assembly of numerous molecules into biological cells does not contravene the law of increasing entropy, the formation of living things does not violate the Second Law of Thermodynamics.

According to the Second Law of Thermodynamics, an enclosed system's entropy will typically rise over time. According to the law, a system has a propensity to become more random or disordered over time. Entropy is a measure of how chaotic or random a system is.

Because biological cells do not form in a closed system, the formation of these cells from a large number of molecules does not violate the Second Law. Since living things are open systems that exchange matter and energy with their surroundings, the Second Law does not apply to them.

In actuality, entropy increases in the environment are caused by living things. Animals, for instance, eat food and excrete waste, which spreads matter and energy around the environment and increases entropy. Entropy is increased in the environment more by living things' metabolic processes than it is reduced by the organization of molecules into cells.

Additionally, biological cell organization is not a natural process. According to the Second Law of Thermodynamics, it necessitates the application of energy and work. Since molecules have a propensity to become disordered by nature, cells rely on energy from their environment to create and maintain their organization.

Because the assembly of molecules into biological cells is not a closed system and requires an input of energy and work, the formation of living organisms does not violate the Second Law of Thermodynamics. Entropy in the environment is increased more by living things' metabolic processes than it is reduced by the organization of their molecules into cells.

3A.1 In an exothermic reaction, 460 J of heat is transferred from the reaction vessel to a large bath of water. Calculate the entropy change of the water if the temperature is (a) 273 K and (b) 373 K.

The entropy change of the water can be calculated using the formula:

S = q / T

where S is the entropy change, q is the heat transferred, and T is the temperature.

(a) At 273 K, the entropy change of the water can be calculated as:

S = 460 J / 273 K

S = 1.68 J/K

Therefore, the entropy change of the water at 273 K is 1.68 J/K.

(b) At 373 K, the entropy change of the water can be calculated as:

S = 460 J / 373 K

S = 1.23 J/K

Therefore, the entropy change of the water at 373 K is 1.23 J/K.

3A.3 A heat engine is required to work at 80 per cent efficiency. At what temperature must the hot source operate if the temperature of the cold sink is 20 C?

The maximum theoretical efficiency of a heat engine operating between two heat reservoirs at temperatures T_h (the temperature of the hot reservoir) and T_c (the temperature of the cold reservoir) is given by the Carnot efficiency:

= 1 - T_c / T_h

where is the efficiency expressed as a fraction between 0 and 1.

In this case, the heat engine is required to operate at 80% efficiency, which means that = 0.8. The temperature of the cold sink is T_c = 20 C, which is equivalent to 293 K (in Kelvin). We need to find the temperature of the hot source, T_h.

Solving the Carnot efficiency equation for T_h, we get:

T_h = T_c / (1 - )

Substituting the values we have, we get:

T_h = 293 K / (1 - 0.8)

T_h = 1465 K

Therefore, the hot source must operate at a temperature of 1465 K, or 1192 C (in Celsius), for the heat engine to operate at 80% efficiency.

3B.4 Calculate the change in entropy of 1.0 kg of lead when it cools from 500 C to 100 C. Use Cp,m = 26.44 J K^1 mol&1.

To calculate the change in entropy of the lead, we can use the formula:

S = (Cp,m/T)dT

where S is the change in entropy, Cp,m is the molar heat capacity at constant pressure, and T is the temperature. We can integrate this expression between the initial and final temperatures of the process.

To use this formula, we need to convert the mass of lead into moles. The molar mass of lead is 207.2 g/mol, so 1.0 kg of lead is equivalent to 1000 g / 207.2 g/mol = 4.824 mol of lead.

Using the given value of Cp,m = 26.44 J K^1 mol^1, we can calculate the change in entropy as:

S = (Cp,m/T)dT from T1 = 500 C = 773 K to T2 = 100 C = 373 K

S = Cp,m ln(T2/T1)

S = (26.44 J K^1 mol^1) * ln(373 K / 773 K) * 4.824 mol

S = -44.4 J/K

Therefore, the change in entropy of 1.0 kg of lead when it cools from 500 C to 100 C is -44.4 J/K. The negative sign indicates that the entropy of the lead has decreased, which is consistent with the fact that cooling is a process in which the system becomes more ordered.

3B.9 Octane is typical of the components of gasoline. Estimate (a) the entropy of vaporization, (b) the enthalpy of vaporization of octane, which boils at 126 C at 1 atm

(a) To estimate the entropy of vaporization of octane, we can use the Clausius-Clapeyron equation:

S = H_vap / T_boil

where S is the entropy of vaporization, H_vap is the enthalpy of vaporization, and T_boil is the boiling temperature.

We can find the boiling temperature of octane at 1 atm using reference data. According to the NIST Chemistry WebBook, the boiling temperature of octane is 125.65 C at 1 atm.

Substituting the values we have, we get:

S = H_vap / T_boil

S = H_vap / (125.65 + 273.15) K

S = H_vap / 398.8 K

To estimate the value of S, we can use a rule of thumb that states that the entropy of vaporization of an organic compound is typically around 85 J/molK. This is not a precise value, but it can be used as an estimate. The molar mass of octane is 114.23 g/mol.

Substituting these values, we get:

85 J/molK = H_vap / 398.8 K

H_vap = 85 J/molK * 398.8 K = 33,908 J/mol

Therefore, the estimated enthalpy of vaporization of octane is 33,908 J/mol.

Note that the above calculation assumes that the entropy of octane in the liquid state is similar to that of a typical organic liquid at room temperature, and that the entropy of octane in the gas state is similar to that of an ideal gas. These assumptions are not always accurate, and a more precise estimate of the entropy of vaporization would require a more detailed thermodynamic analysis.

3C.3 An FClO3 molecule can adopt four orientations in the solid with negligible difference in energy. What is its residual molar entropy?

S = R lnW

where R is the gas constant and W is the number of available microstates. For a molecule that can adopt four orientations with negligible difference in energy, there are four equivalent microstates, each of which corresponds to one of the four orientations. Therefore, the number of microstates is W = 4.

Substituting R = 8.314 J/molK and W = 4, we get:

S = R lnW

S = (8.314 J/molK) * ln(4)

S = 11.6 J/molK

Therefore, the residual molar entropy of an FClO3 molecule that can adopt four orientations in the solid state is 11.6 J/molK. This value is relatively small compared to the molar entropies of most substances at room temperature, which typically range from a few tens to a few hundreds of J/molK.

.2 Use the information on standard molar entropies in the Resource section to calculate the standard reaction entropy at 298 K of

(a) 2 CH3CHO(g) + O2(g) 2 CH3COOH(l)

(b) 2 AgCl(s) + Br2(l) 2 AgBr(s) + Cl2(g)

(d) Zn(s) + Cu^2+ (aq) Zn^2+ (aq) + Cu(s)

(e) C12H22O11(s) + 12 O2(g) 12 CO2(g) + 11 H2O(l)

The standard reaction entropy at 298 K can be calculated using the standard molar entropies of the reactants and products. The standard reaction entropy is given by:

S = nS(products) - mS(reactants)

where n and m are the stoichiometric coefficients of the products and reactants, respectively, and S is the standard molar entropy.

(a) 2 CH3CHO(g) + O2(g) 2 CH3COOH(l)