Consider the apparatus shown below, in which two 1.00-L bulbs are initially sepa
ID: 711164 • Letter: C
Question
Consider the apparatus shown below, in which two 1.00-L bulbs are initially separated from one another by a other with 400. torr of oxygen, and the entire apparatus is at an initial temperature of 30. c closed valve. One bulb is filled with 501 torr of sulfur dioxide and the Before mixing 02 V-1.00L SO2 V1.00L P-5.01 × 102 torr P-4.00 × 102 torr T=30°C T-30 C The valve separating the bulbs is opened so that the gases mix, the entire container is heated, and the two gases react to produce the maximum possible amount of SOs. If the container is at a final temperature of 616 K when the reaction reaches completion, what is the total pressure? Pressure- torr Bubmit Answer Try Anothar Version 3 Item attempts remainingExplanation / Answer
According to ideal gas equation PV = nRT
P = preesure of the gas
V = volume of the gas
n = number of oles of the gas
R = universal gas constant
T = Absolute temperature
Number of moles of SO2 n = PV/RT = (501/760)*1/(0.0821*303) = 0.0265 moles
number of moles of O2 n = PV/RT = (400/760)*1/(0.0821*303) = 0.02116 moles
After valve opening the two gases react with each other and volume of the mixture is additive of two bulbs,
2SO2 + O2 -------> 2SO3
2 moles of SO2 produces 2 moles of SO3
0.0265 moles of SO2 produces 0.0265 moles of SO3
1 mole of O2 produces 2 moles of SO3
0.02116 moles of O2 produces 2*0.02116 moles of SO3
= 0.04232 molesa of SO3
so SO2 is limiting reagent for the completion of the reaction.
2 moles of SO3 require 1 mole of O2
0.0265 moles of SO3 require 1/2*0.0265 moles of O2
= 0.01325 moles of O2
remaining moles of O2 = 0.02116 - 0.01325 = 0.00791 moles
Total Pressure = Partial pressure of SO3 + Partial pressure of O2
Pt = nRT/V for SO3 + nRT/V for O2
Pt = (0.0265*0.0821*616/2) + (0.00791*0.0821*616/2)
Pt = 0.8701 atm
Pt = 0.8701 * 760 = 661.276 torrs