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Studentka2010 [4]
3 years ago
15

Which response includes all of the following processes that are accompanied by an increase in entropy?

Chemistry
1 answer:
dlinn [17]3 years ago
7 0

Answer:

1 and 3.

Explanation:

The entropy measures the randomness of the system, as higher is it, as higher is the entropy. The randomness is associated with the movement and the arrangement of the molecules. Thus, if the molecules are moving faster and are more disorganized, the randomness is greater.

So, the entropy (S) of the phases increases by:

S solid < S liquid < S gases.

1. The substance is going from solid to gas, thus the entropy is increasing.

2. The substance is going from a disorganized way (the molecules of I are disorganized) to an organized way (the molecules join together to form I2), thus the entropy is decreasing.

3. The molecules go from an organized way (the atom are joined together) to a disorganized way, thus the entropy increases.

4. The ions are disorganized and react to form a more organized molecule, thus the entropy decreases.

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
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<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

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