Be sure to answer all parts.The equilibrium constant (K) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl½(g)= 2NOCI(g) is 5 x 10° at a certain temperature. In an experiment, 2.10 × 10 ² mole of NO, 2.10 × 103 mole of Cl,, and 2.80 moles of NOC1 are mixed in a 2.20–L flask. What is Q, for the experiment? x 10 (Enter your answer in scientific notation.) In which direction will the system proceed to reach equilibrium? The reaction will proceed to the left. The reaction will proceed to the right. The reaction is at equilibrium.
Be sure to answer all parts.The equilibrium constant (K) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl½(g)= 2NOCI(g) is 5 x 10° at a certain temperature. In an experiment, 2.10 × 10 ² mole of NO, 2.10 × 103 mole of Cl,, and 2.80 moles of NOC1 are mixed in a 2.20–L flask. What is Q, for the experiment? x 10 (Enter your answer in scientific notation.) In which direction will the system proceed to reach equilibrium? The reaction will proceed to the left. The reaction will proceed to the right. The reaction is at equilibrium.
World of Chemistry, 3rd edition
3rd Edition
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Chapter17: Equilibrium
Section: Chapter Questions
Problem 38A
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Transcribed Image Text:Be sure to answer all parts.The equilibrium constant (K) for the formation of nitrosyl chloride, an
orange-yellow compound, from nitric oxide and molecular chlorine
2NO(g) + Cl,(g)= 2NOCI(g)
is 5 x 10° at a certain temperature. In an experiment, 2.10 × 10-2 mole of NO, 2.10 x 10-3 mole of Cl,
and 2.80 moles of NOCI are mixed in a 2.20-L flask.
What is Q, for the experiment?
× 10
(Enter your answer in scientific notation.)
In which direction will the system proceed to reach equilibrium?
The reaction will proceed to the left.
The reaction will proceed to the right.
The reaction is at equilibrium.
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