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To determine the enthalpy change (ΔHΔH) for the reaction H2(g)+Br2(g)→2HBr(g)H2(g)+Br2(g)→2HBr(g), we need to calculate the energy required to break all the bonds in the reactant molecules and subtract the energy required to form all the bonds in the product molecules.

Given bond energies:

• Bond energy of H2H2 (H-H) = 4.35 kJ/mol
• Bond energy of Br2Br2 (Br-Br) = 192 kJ/mol
• Bond energy of HBrHBr (H-Br) = 368 kJ/mol

The net change in enthalpy (ΔHΔH) can be calculated using the following formula:

ΔH=∑(bond energies of reactants)−∑(bond energies of products)ΔH=∑(bond energies of reactants)−∑(bond energies of products)

For the given reaction:

• Reactants: H2(g)+Br2(g)H2(g)+Br2(g)
• Products: 2HBr(g)2HBr(g)

Let's calculate:

1. Energy required to break bonds in reactants: H2(g)H2(g): 1 mole of H2H2 contains 1 bond, so energy required = 1×Bond energy of H21×Bond energy of H2 Br2(g)Br2(g): 1 mole of Br2Br2 contains 1 bond, so energy required = 1×Bond energy of Br21×Bond energy of Br2

   Total energy required to break bonds in reactants = Bond energy of H2+Bond energy of Br2Bond energy of H2+Bond energy of Br2

2. Energy released when bonds are formed in products: 2HBr(g)2HBr(g): 1 mole of 2HBr2HBr contains 2 bonds, so energy released = 2×Bond energy of HBr2×Bond energy of HBr

   Total energy released when bonds are formed in products = 2×Bond energy of HBr2×Bond energy of HBr

Now, let's substitute the given bond energies and calculate ΔHΔH:

ΔH=(Bond energy of H2+Bond energy of Br2)−(2×Bond energy of HBr)ΔH=(Bond energy of H2+Bond energy of Br2)−(2×Bond energy of HBr)

ΔH=(4.35 kJ/mol+192 kJ/mol)−(2×368 kJ/mol)ΔH=(4.35 kJ/mol+192 kJ/mol)−(2×368 kJ/mol)

ΔH=(196.35 kJ/mol)−(736 kJ/mol)ΔH=(196.35 kJ/mol)−(736 kJ/mol)

ΔH=−539.65 kJ/molΔH=−539.65 kJ/mol

Therefore, the enthalpy change (ΔHΔH) for the given reaction is -539.65 kJ/mol.

The difference between the molar specific heats at constant pressure (CpCp) and constant volume (CvCv) for an ideal gas can be derived using the relationship between enthalpy (HH), internal energy (UU), pressure (pp), and volume (VV):

H=U+pVH=U+pV

For an ideal gas, the internal energy only depends on temperature and the number of moles (nn):

U=nCvTU=nCvT

And, for an ideal gas, the enthalpy depends on the internal energy, pressure, and volume:

H=nCpTH=nCpT

Using these relationships and the given empirical relation, we can derive the difference between CpCp and CvCv:

H=U+pVH=U+pV

nCpT=nCvT+pnRTnCpT=nCvT+pnRT

Dividing both sides by nTnT:

Cp=Cv+pVnTCp=Cv+nTpV

For an ideal gas, we know that pV=nRTpV=nRT, so:

Cp=Cv+nRTnTCp=Cv+nTnRT

Cp=Cv+RCp=Cv+R

Therefore, the difference between CpCp and CvCv for an ideal gas is equal to the universal gas constant (RR).

Given that we're considering 10 moles of an ideal gas, we'll use the value of the universal gas constant (RR) for the calculation. The value of RR depends on the units used, but commonly it's approximately 8.314 J/mol⋅K8.314J/mol⋅K.

Difference between Cp and Cv=RDifference between Cp and Cv=R

Difference between Cp and Cv=10 moles×8.314 J/mol⋅KDifference between Cp and Cv=10moles×8.314J/mol⋅K

Difference between Cp and Cv≈83.14 J/KDifference between Cp and Cv≈83.14J/K

So, for 10 moles of an ideal gas, the difference between CpCp and CvCv is approximately 83.14 J/K83.14J/K.

The relation between heat capacity (CpCp) and specific heat (cc) for 1 mole of a substance, such as water, can be derived from their definitions.

1. Heat Capacity (CpCp): Heat capacity (CpCp) is an extensive property, meaning it depends on the amount of substance present. It is defined as the amount of heat required to raise the temperature of a given amount of substance by 1 degree Celsius (or 1 Kelvin).

Cp=n×cCp=n×c

Where:

• nn is the number of moles of the substance.
• cc is the specific heat capacity of the substance.

2. Specific Heat (cc): Specific heat (cc) is an intensive property, meaning it does not depend on the amount of substance present. It is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 Kelvin).

c=Cpnc=nCp

For 1 mole of water (n=1 moln=1mol), the relation between CpCp and cc becomes:

c=Cp1=Cpc=1Cp=Cp

Therefore, for 1 mole of water, the specific heat (cc) is equal to the heat capacity (CpCp). In other words, the specific heat of 1 mole of water is numerically equal to its heat capacity.

When an ideal gas expands isothermally into a vacuum, it undergoes free expansion. In free expansion, no external pressure is exerted on the gas, so the gas performs no work on its surroundings. Additionally, since the temperature remains constant during isothermal expansion, the change in internal energy (ΔUΔU) of the gas is also zero.

Therefore, in free expansion:

1. Work done (WW): Since there is no external pressure (pext=0pext=0) acting on the gas, the work done by the gas (WW) is zero.

W=0W=0

2. Change in internal energy (ΔUΔU): Since the process is isothermal (ΔT=0ΔT=0), the change in internal energy (ΔUΔU) is also zero.

ΔU=0ΔU=0

So, in this case, for the expansion of 1 litre of ideal gas into vacuum until its total volume is 5 litres:

• Work done (WW) = 0
• Change in internal energy (ΔUΔU) = 0

Under certain specific conditions, the heat absorbed or released by a system can be independent of the path taken during the process. These conditions are associated with two types of thermodynamic processes:

1. Isochoric Process (Constant Volume): In an isochoric process, the volume of the system remains constant. Since there is no change in volume (ΔV=0ΔV=0), the work done by or on the system is zero (W=0W=0). According to the first law of thermodynamics:

ΔU=Q−WΔU=Q−W

Since W=0W=0 in an isochoric process, the change in internal energy (ΔUΔU) is equal to the heat added to or removed from the system (QQ). Therefore, the heat absorbed or released by the system in an isochoric process is independent of the path taken.

2. Adiabatic Process: In an adiabatic process, no heat is exchanged between the system and its surroundings (Q=0Q=0). Therefore, according to the first law of thermodynamics:

ΔU=Q−WΔU=Q−W

With Q=0Q=0, the change in internal energy (ΔUΔU) is equal to the work done on or by the system (WW). The magnitude and direction of the work depend solely on the initial and final states of the system and are independent of the path taken. Therefore, the heat absorbed or released by the system in an adiabatic process is also independent of the path taken.

In summary, the heat absorbed or released by a system is independent of the path taken under the following specific conditions:

• Isochoric process (constant volume), where ΔV=0ΔV=0
• Adiabatic process, where Q=0Q=0

At equilibrium, the Gibbs free energy change (ΔrGΔrG) will be zero. This is because at equilibrium, the forward and reverse reactions occur at the same rate, and there is no net change in the system.

The standard Gibbs free energy change (ΔrG∘ΔrG∘) is the Gibbs free energy change for a reaction under standard conditions (usually at 1 atm pressure and a specified temperature, often 298 K). ΔrG∘ΔrG∘ gives the Gibbs free energy change when all reactants and products are present at standard states (usually 1 M concentration for solutions).

While ΔrG∘ΔrG∘ provides important information about the thermodynamic feasibility of a reaction, it does not necessarily indicate whether a reaction is at equilibrium. Only when ΔrGΔrG is zero does it indicate that the reaction has reached equilibrium.

The standard molar entropy of a substance typically increases when it undergoes a phase change from a solid to a liquid, and then from a liquid to a gas. This is because as a substance transitions from a more ordered state (solid) to a less ordered state (liquid or gas), the number of possible microstates increases, leading to an increase in entropy.

Therefore, the standard molar entropy of H2O(s)H2O(s) is expected to be greater than that of H2O(l)H2O(l).

This is because in the solid phase, the molecules of water (H2OH2O) are more ordered and constrained in their positions, resulting in lower entropy compared to the liquid phase, where the molecules are free to move and have more disorder.

So, the standard molar entropy of H2O(s)H2O(s) is expected to be more than 70 J K−1−1 mol−1−1.

The mathematical relation that relates heat (qq), temperature (TT), and entropy (SS) is given by the following equation:

q=T⋅ΔSq=T⋅ΔS

This equation represents the change in entropy of a system (ΔSΔS) due to the transfer of heat (qq) at a constant temperature (TT). It shows how the transfer of heat influences the disorder or randomness of a system, which is described by its entropy.

In words, this equation can be interpreted as follows: The heat transferred to or from a system at constant temperature is directly proportional to the change in entropy of the system.

This relation highlights the connection between heat, temperature, and entropy, demonstrating that the transfer of heat can lead to changes in the disorder or randomness of a system, which is reflected in its entropy.