Question

A chemist studying the pollutant SO_2 (g) places 1.00 mole(s) of SO_2 (g) and 1.00 mole(s) of O_2 in a flexible container at T = 509 degree C and 1.69 atm. And some gaseous SO_3 was produced. What is the final volume (in L) of the container after the reaction has gone to completion? Is work done on the system or by the system? (You only have one try for this question) Done by the system Done on the system

Answer 1

a) reaction is

2SO₂ + O₂   $\rightarrow$ 2SO₃

According to reaction 2 mole of SO₂ react with 1 mole of O₂ then for complete reaction 1 mole of SO₂ require 0.5 mole of O₂ thus, 0.5 mole of O₂ remain unreacted

and 2 mole of SO₂ produce 2 mole of SO₃ then 1 mole of SO₂ produce 1 mole of SO₃

total mole of container after complete reaction = 0.5 mole O₂ + 1 mole SO₃ = 1.5 mole

to calculate final volume use ideal gas equation .

We know that PV = nRT

V = nRT/P

n = 1.5 mole,

T = 509⁰C = 509+273.15 = 782.15K,

P= 1.69 atm,

R = 0.08205 L atm mol^-1 K^-1 ( R = gas constant)

V = ?

Substitute these value in above equation.

V = 1.5 $\times$ 0.08205 $\times$ 782.15/1.69 = 56.96 L

final volume of gas is 56.96 L

b) calculate initial volume

initial mole = 1 mole SO₂ + 1 mole O₂ = 2 mole

to calculate initial volume use ideal gas equation .

We know that PV = nRT

V = nRT/P

n = 2 mole,

T = 509⁰C = 509+273.15 = 782.15K,

P= 1.69 atm,

R = 0.08205 L atm mol^-1 K^-1 ( R = gas constant)

V = ?

Substitute these value in above equation.

V = 2 $\times$  0.08205  $\times$ 782.15/1.69 = 75.94 L

initial volume of gas is 75.94 L

nitial volume of gas is 75.94 L and final volume of gas is 56.96 L thus compession take place

therefore work done on the system.