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

Step-by-step explanation:

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

`p_A=p_T* \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=(n_A)/(n_A+n_B)`

`\chi_(SO_3)=(17.0)/(26.0+83.0+17.0)=0.135`

Putting the values in equation (1):

`p_(SO_3)=p_T* \chi_(SO_3)`

`p_(SO_3)=0.950atm* 0.135=0.128atm`

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