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Leya [2.2K]
2 years ago
15

Consider a closed containing a solid in equilibrium with its vapor. The volume of the solid is much less than that of the contai

ner. Let the partition function of the solid be given by Q-qss where Ns is the number of molecules in the solid phase. Minimize the total Helmholtz free energy at fixed N,V, and T, and show that the equilibrium condition is given by Ng=qo/q2 where Ng is the number of gas phase molecules and qg is the partition function of a gas phase molecule.

Chemistry
1 answer:
Furkat [3]2 years ago
5 0

Answer:

Explanation:

check the attachment below

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2 years ago
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What is the mass of a sample of alcohol (specific heat = 2.4 J/gC), if it requires 4780 J of heat to raise the temperature by 5.
insens350 [35]

The mass of a sample of alcohol is found to be = m = 367 g

Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.

Explanation:

The Energy of the sample given is q = 4780

We are required to find the mass of alcohol m = ?

Given that,

The specific heat given is represented by = c = 2.4 J/gC

The temperature given is ΔT = 5.43° C

The mass of sample of alcohol can be found as follows,

The formula is c = \frac{q}{mt}

We can drive value of m bu shifting m on the left hand side,

m = \frac{q}{ct}

mass of alcohol (m) = \frac{4780}{(2.4)( 5.43)}

m = 367 g

Therefore, The mass of the given sample of alcohol is

m = 367g

It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.

4 0
3 years ago
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

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Explanation:

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