Suppose a 500. mL flask is filled with 1.3 mol of O, and 0.10 mol of SO3. This reaction becomes possible: 250,(g) + 0,(g) = 2so;(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of O,. You can leave out the M symbol for molarity. so, 0, so,
Suppose a 500. mL flask is filled with 1.3 mol of O, and 0.10 mol of SO3. This reaction becomes possible: 250,(g) + 0,(g) = 2so;(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of O,. You can leave out the M symbol for molarity. so, 0, so,
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Transcribed Image Text:Suppose a 500. mL flask is filled with 1.3 mol of O, and 0.10 mol of SO3. This reaction becomes possible:
2s0,(g) +0,(g) – 2s0;(g)
Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the
equilibrium molarity of each compound after the reaction has come to equilibrium.
Use x to stand for the unknown change in the molarity of 0,. You can leave out the M symbol for molarity.
2.
so,
02
so,
initial
change
equilibrium
Expert Solution
Step 1
GIVEN:
Volume of the Flask = 500 mL= 0.5 L
Moles of O2=1.3 mol
Moles of SO3 =0.10 mol
Given reaction is:
2SO2(g) + O2 <------> 2SO3(g)
Step 2
Concentration = Moles/ Volume (L)
Concnetration of O2 = 1.3 mol / 0.5 L
= 2.6 M
Concentration of SO3= 0.10 mol/0.5 L
= 0.2 M
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