Question

A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

What is the initial effect of the change on vapour pressure?

How do rates of evaporation and condensation change initially?

What happens when equilibrium is restored finally and what will be the final vapour pressure?

Assign oxidation numbers to the underlined elements in each of the following species:
$NaH\underset{–}{S}{O}_4$

$H_4{\underset{–}{P}}_2{O}_7$

$K_2\underset{–}{Mn}{O}_4$

$Ca{\underset{–}{O}}_2$

$Na\underset{–}{B}{H}_4$

$H_2{\underset{–}{S}}_2{O}_7$

$KAl\underset{–}{S}{O}_4{)}_2.12H_{2}O$

Answer 1

If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.

Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases.
As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.

When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume.

Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

$NaH\underset{–}{S}{O}_4$
Then, we have
$1(+1)+1(+1)+1\left(x\right)+4(-2)=0\Rightarrow 1+1+x-8=0\Rightarrow x=+6$
Hence, the oxidation number of S is + 6.

$H_4{\underset{–}{P}}_2{O}_7$
$\stackrel{+1}{H_4}\stackrel{x-2}{P_2{O}_7}$
Then, we have
$4(+1)+2\left(x\right)+7(-2)=0\Rightarrow 4+2x-14=0\Rightarrow 2x=+10\Rightarrow x=+5$
Hence, the oxidation number of P is + 5.

$K_2\underset{–}{Mn}{O}_4$
$\stackrel{+1}{K_2}\stackrel{x-2}{Mn{O}_4}$
Then, we have
$2(+1)+x+4(-2)=0\Rightarrow 2+x-8=0\Rightarrow x=+6$
Hence, the oxidation number of Mn is + 6.

$H_2{\underset{–}{S}}_2{O}_7$
$\stackrel{+1}{H_2}\stackrel{x-2}{S_2{O}_7}$
Then, we have
$2(+1)+2\left(x\right)+7(-2)=0\Rightarrow 2+2x-14=0\Rightarrow 2x=12\Rightarrow x=+6$
Hence, the oxidation number of S is + 6.

$Ca{\underset{–}{O}}_2$
Then, we have
$\stackrel{+2}{C_a}\stackrel{x}{O_2}(+2)+2\left(x\right)=0\Rightarrow 2+2x=0\Rightarrow x=-1$
Hence, the oxidation number of O is -1.

$Na\underset{–}{B}{H}_4$
$\stackrel{+1}{Na}\stackrel{x-1}{B{H}_4}$
Then, we have
$1(+1)+1\left(x\right)+4(-1)=0\Rightarrow 1+x-4=0\Rightarrow x=+3$
Hence, the oxidation number of B is + 3.

$KA1(\underset{–}{S}{O}_4{)}_2.12H_{2}O\stackrel{+13+}{KA1}{\left(\stackrel{x2-}{S{O}_4}\right)}_2.12\stackrel{+1-2}{H_{2}O}$
Then, we have
$1(+1)+1(+3)+2\left(x\right)+8(-2)+24(+1)+12(-2)=0\Rightarrow 1+3+2x-16+24-24=0\Rightarrow 2x=12\Rightarrow x=+6$
Or,
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have
$1(+1)+1(+3)+2\left(x\right)+8(-2)=0\Rightarrow 1+3+2x-16=0\Rightarrow 2x=12\Rightarrow x=+6$
Hence, the oxidation number of S is + 6.