Question

Sulfur dioxide reacts with oxygen in the presence of plati- num to give sulfur trioxide: 2 SO2(g) " O2(g) 8n 2 SO3(g) Suppose that at one stage in the reaction, 26.0 mol SO2, 83.0 mol O2, and 17.0 mol SO3 are present in the reaction vessel at a total pressure of 0.950 atm. Calculate the mole fraction of SO3 and its partial pressure.

Answer 1

Answer: mole fraction of $SO_3$ = 0.135

partial pressure of $SO_3$ = 0.128 atm

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

$p_A=p_T\times \chi_A$ ......(1)

where,

$p_A$ = partial pressure of substance A

$p_T$ = total pressure

$\chi_A$ = mole fraction of substance A

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{SO_3}=\frac{17.0}{26.0+83.0+17.0}=0.135$

Putting the values in equation (1):

$p_{SO_3}=p_T\times \chi_{SO_3}$

$p_{SO_3}=0.950atm\times 0.135=0.128atm$

Thus the mole fraction of $SO_3$ and its partial pressure are 0.135 and 0.128 atm.