Chapter 17, Problem 17.1P

Interpretation Introduction

Interpretation: Spontaneous reactions are self driven reactions; they take place immediately without any assistance from external source. The spontaneity of a process depends on several factors; the reaction is spontaneous if   $\Delta S^o{}_{sys}$ is positive or   $\Delta H^o{}_{sys}$ is negative.

Concept introduction: The spontaneity of a reaction can be determined by the equation given below.

If the sign of $\Delta G_{system\text{}}$ is negative the reaction is spontaneous, and if it is positive, the reaction is non-spontaneous.

  $\Delta G_{system\text{}}=\Delta H_{system\text{}}-T\Delta S_{system}\text{}$

Temperature in this equation is taken in Kelvin, it is always positive (or zero).

Given:

The given processes are-

1. Diffusion of perfume molecules from one side of a room to the other side.
2. Heat flow from a cold object to a hot object.
3. Decomposition of rust $(F{e}_2{O}_3. H_{2}O)$ to iron metal, oxygen, and water.
4. A mixture of gaseous $N_2$ , $H_2$ , and $N{H}_3$ , with the partial pressures
$(P_{N_2}=1 atm, P_{H_2}=0.25 atm, P_{N{H}_3}=10 atm),$ converts some of the $N_2$ and $H_2$ to $N{H}_3$ .

  $N_2(g)+3H_2(g)\stackrel{Catalyst}{\rightleftharpoons }2N{H}_3(g)$ at 300 K

  $K_p=4.4\times {10}^5$

To determine:

To determine which of the given processes are spontaneous and which are non-spontaneous.

Answer to Problem 17.1P

Solution:

The given processes are-

1. Spontaneous
2. Non-spontaneous
3. Non-spontaneous
4. Spontaneous

Explanation of Solution

1. Diffusion of perfume molecules from one side of a room to the other side.

This process is governed by change in entropy.

The molecules of perfume, when move from one side of a room to the other side experience increase in the volume where, the particles move randomly. Randomness increases due to the increase in the volume. The change in entropy is positive, therefore the process is spontaneous.

2. Heat flow from a cold object to a hot object.

This process is governed by temperature.

Heat flow from a system to surroundings is an exothermic process and is favourable at low temperature. Since the surrounding is at higher temperature, this reaction is endothermic, and change in enthalpy is positive, therefore the process is non-spontaneous.

3. Decomposition of rust $(F{e}_2{O}_3. H_{2}O)$ to iron metal, oxygen, and water.

This process is reverse of rusting of Iron. Rusting of Iron is a spontaneous process due to the continuous availability of oxygen and moisture from the environment. But the reverse is not spontaneous at all, as there iron that has rusted will never loose moisture and oxygen to for Iron.

4. A mixture of gaseous $N_2$ , $H_2$ , and $N{H}_3$ , with the partial pressures

$(P_{N_2}=1 atm, P_{H_2}=0.25 atm, P_{N{H}_3}=10 atm),$ converts some of the $N_2$ and $H_2$ to $N{H}_3$ .

  $N_2(g)+3H_2(g)\stackrel{Catalyst}{\rightleftharpoons }2N{H}_3(g)$ at 300 K

  $K_p=4.4\times {10}^5$

Calculating reaction quotient, Q

  $Q=\frac{\mathrm{Pr}essure of \mathrm{Pr}oducts}{\mathrm{Pr}essure of \mathrm{Re}ac\tan ts}$

  $Q=\frac{{\left[P_{N{H}_3}\right]}^2}{\left[P_{N_2}\right]{\left[P_{H_2}\right]}^3}$

  $Q=\frac{{\left[10\right]}^2}{\left[1\right]{\left[0.25\right]}^3}=6.4\times {10}^3$

Since, $K_p$ is higher than Q, therefore the reaction is spontaneous.

Conclusion

The given processes are-

1. Spontaneous
2. Non-spontaneous
3. Non-spontaneous
4. Spontaneous