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A student ran the following reaction in the laboratory at 1144 K: 2S02(g) + O2(g) = 2503(g) When she introduced 8.87×10-2 moles of SO2(g) and 7.81×10-2 moles of O2(g) into a 1.00 liter container, she found the equilibrium concentration of SO3(g) to be 5.19×10-2 M. Calculate the equilibrium constant, K., she obtained for this reaction. K. =

A student ran the following reaction in the laboratory at 1144 K: 2S02(g) + O2(g) = 2503(g) When she introduced 8.87×10-2 moles of SO2(g) and 7.81×10-2 moles of O2(g) into a 1.00 liter container, she found the equilibrium concentration of SO3(g) to be 5.19×10-2 M. Calculate the equilibrium constant, K., she obtained for this reaction. K. =

Chemistry
Chemistry
10th Edition
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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A student ran the following reaction in the laboratory at 1144 K: 2502(g) + O2(g) = 2503(g) When she introduced 8.87x102 moles of SO2(g) and 7.81×102 moles of O2(g) into a 1.00 liter container, she found the equilibrium concentration of SO3(g) to be 5.19x10-2 M. Calculate the equilibrium constant, K., she obtained for this reaction. K.
Transcribed Image Text:A student ran the following reaction in the laboratory at 1144 K: 2502(g) + O2(g) = 2503(g) When she introduced 8.87x102 moles of SO2(g) and 7.81×102 moles of O2(g) into a 1.00 liter container, she found the equilibrium concentration of SO3(g) to be 5.19x10-2 M. Calculate the equilibrium constant, K., she obtained for this reaction. K.
Expert Solution
Step 1

Given:

The reaction is as follows:

2SO2(g)         +         O2(g)        ⇌            2SO3(g).

The initial moles of SO2(g)  = 8.87×10-2 moles=0.0887 moles.

The initial moles of O2(g)  = 7.81×10-2 moles=0.0718 moles.

The volume of the container, (V) = 1 L.

NOTE-The moles and the concentration of the given species will be equal because the volume of the container is 1 L.

Therefore, the given moles of SO3 at equilibrium = 5.19×10-2×1 L=0.0519 moles

 

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