At an elevated temperature, K, = 4.0 × 10-6 for the reaction 2 HBr(g) = H2(g) + Br2(g). If the initial partial pressures of HBr, H2, and Br2 are 1.0 x 10-2 atm, 2.0 × 10- 4 atm, and 2.0 × 10-4 atm, respectively, what is the equilibrium partial pressure of H2?
At an elevated temperature, K, = 4.0 × 10-6 for the reaction 2 HBr(g) = H2(g) + Br2(g). If the initial partial pressures of HBr, H2, and Br2 are 1.0 x 10-2 atm, 2.0 × 10- 4 atm, and 2.0 × 10-4 atm, respectively, what is the equilibrium partial pressure of H2?
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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
Transcribed Image Text:At an elevated temperature, K, = 4.0 × 10-6 for the reaction 2 HBr(g) = H2(g) +
Br2(g). If the initial partial pressures of HBr, H2, and Br2 are 1.0 × 10-2 atm, 2.0 × 10-
3.
4
atm, and 2.0 × 10-4 atm, respectively, what is the equilibrium partial pressure of
6.
H2?
At 1000 K, Kp = 19.9 for the reaction Fe203(s) + 3 CO(g) = 2 Fe(s) + 3 CO2(g). What
is the value of K, for the reaction 4 Fe203(s) + 12 CO(g) = 8 Fe(s) + 12 CO2(g)?
Exactly 3.75 moles of N204 is placed in an empty 1.0-L container. After a period of
time, equilibrium is reached and described by the equation below:
N204 (g) = 2NO2 (g)
If at equilibrium the N204 is 45% dissociated, what is the value of the equilibrium
constant for the reaction?
![What is the equilibrium equation for the reaction:
3
NH4NO3(s) = N20(g) + 2 H2O(/)?
O A) Ke = [N2O][H2O12
%3D
6.
5-
B) Kc = [N20][H2O]
8.
9.
--
N20]H20]2
Kc
%3D
NHẠNO3]
O D) K = [N20]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff27c3bcc-31ef-40b0-b008-4b23339d198c%2F242b8ba2-0691-4047-8c33-7ab2ca92efb6%2Fjj9ssgaw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:What is the equilibrium equation for the reaction:
3
NH4NO3(s) = N20(g) + 2 H2O(/)?
O A) Ke = [N2O][H2O12
%3D
6.
5-
B) Kc = [N20][H2O]
8.
9.
--
N20]H20]2
Kc
%3D
NHẠNO3]
O D) K = [N20]
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