Question

Consider the following changes:

$\begin{array}{l}\mathbf{a}\mathbf{.}{\mathbf{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\\ \mathbf{b}\mathbf{.}\mathbf{C}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{c}\mathbf{.}\mathbf{C}{\mathbf{a}}_{\mathbf{3}}{\mathbf{P}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{3}\mathbf{C}\mathbf{a}{\mathbf{\left(}\mathbf{O}\mathbf{H}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{P}{\mathbf{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{d}\mathbf{.}\mathbf{2}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{O}\mathbf{H}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\\ \mathbf{e}\mathbf{.}{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\end{array}$

At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Short answer

a.the surrounding will do compression work on the system.

b. nowork is done on/by the system.

c. work is done by the system on the surroundings.

d. the surrounding will do compression work on the system.

e. work is done by the system on the surroundings.

Step-by-step solution

Step 1:Under these conditions, when temperature and pressure both are constant,

We generally use the formula:

$w=-P\Delta V$

The convention is that if w is negative, work is done by the system and if w is positive, work is done on the system.

Now write down the balanced chemical reaction for all the above chemical processes and find out the change in the number of moles (n) that occurs during the chemical reaction:

So, for the reactions that are given, we need to find out if the volume increases or decreases, or stays the same (P is constant)

From,$w=-P\Delta V$

Δn = moles of gaseous products − moles of gaseous reactants.

Only gases can do PV work (we ignore solids and liquids). When a balanced reaction has more moles of product gases than moles of reactant gases

(Δn positive), the reaction will expand in volume (ΔV positive), and the system will do work on the surroundings.

a.$N_2\left(g\right)\to N_2\left(l\right)$;(Δn = 0 – 1) = -1

since moles of gases decrease from reactant to products the reaction will contract in volume(∆V negative)

Hence the surrounding will do compression work on the system.

$b.CO\left(g\right)+H_{2}O\left(g\right)\to H_2\left(g\right)+C{O}_2\left(g\right)$;(Δn = 2 – 2) = 0

since moles of gases remain the same from reactant to products therefore no change is occurring in volume(∆V = 0)

Hence nowork is done on/by the system.

$c.C{a}_3{P}_2\left(s\right)+6H_{2}O\left(l\right)\to 3Ca{\left(OH\right)}_2\left(s\right)+2P{H}_3\left(g\right)$; (Δn = 2-0 ) = 2

since moles of gases decrease from reactant to products the reaction will contract in volume (∆V positive)

Hencework is done by the system on the surroundings.

$d.2C{H}_{3}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+4H_{2}O\left(l\right)$; (Δn = 2-3) = -1

since moles of gases decrease from reactant to products the reaction will contract in volume(∆V negative)

Hence the surrounding will do compression work on the system.

$e.I_2\left(s\right)\to I_2\left(g\right)$; (Δn = 1-0 ) = 1

since moles of gases decrease from reactant to products the reaction will contract in volume(∆V positive)

Hence work is done by the system on the surroundings.