Question

Sulphur dioxide and oxygen react according to the equation :- $2S{O}_2\left(g\right)+O_2\left(g\right)\to 2S{O}_3\left(g\right)$.If 50.0 g of $S{O}_2$(g) is added to a flask resulting in a pressure of 0.750 atm.What will be the total pressure in the flask when a stochiometric amount of oxygen is added?What is the pressure after the completion of reaction and before the beginning of the reaction (with $O_2$ just added)?

• A. Before reaction = 1.125 atm , After reaction = 0.725 atm
• B. Before reaction = 2.50 atm , After reaction = 1.000 atm
• C. Before reaction = 1.125 atm , After reaction = 0.750 atm
• D. Before reaction = 2.50 atm , After reaction = 1.125 atm

Answer 1

The correct option is C

Before reaction = 1.125 atm , After reaction = 0.750 atm

Since the stochiometric ratio of the $S{O}_{2}andS{O}_3$ are the same,the pressure in the flask after the reaction when the only substance in the flask is$S{O}_3$ will be the same as the pressure in flask when there is just $S{O}_2$ is 0.750 atm.

From the equation , stochiometric ratio of the $O_2$ is half of that of $S{O}_2$ .So after oxygen is added, the total pressure in the flask will be $0.750+\frac{0.750}{2}=1.125atm$.So,the pressure will be 1.125 atm when the $O_2$ is just added.