At high temperatures nitrogen (N2) and oxygen (O2) will react to form NO. N2(g) + O2(g) ⇆ 2 NO(g) (4.1) The value for the equilibrium constant for reaction 4.1 is KC = 2.7 x 10^-17 at some temperature T. A system initially has [N2] = 0.0800 M and [O2] = 0.0500 M. There is no NO initially present in the system. Find the value for [NO] that will be present at equilibrium.
At high temperatures nitrogen (N2) and oxygen (O2) will react to form NO. N2(g) + O2(g) ⇆ 2 NO(g) (4.1) The value for the equilibrium constant for reaction 4.1 is KC = 2.7 x 10^-17 at some temperature T. A system initially has [N2] = 0.0800 M and [O2] = 0.0500 M. There is no NO initially present in the system. Find the value for [NO] that will be present at equilibrium.
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.78PAE
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At high temperatures nitrogen (N2) and oxygen (O2) will react to form NO.
N2(g) + O2(g) ⇆ 2 NO(g) (4.1)
The value for the equilibrium constant for reaction 4.1 is KC = 2.7 x 10^-17 at some temperature T.
A system initially has [N2] = 0.0800 M and [O2] = 0.0500 M. There is no NO initially present in
the system. Find the value for [NO] that will be present at equilibrium.
N2(g) + O2(g) ⇆ 2 NO(g) (4.1)
The value for the equilibrium constant for reaction 4.1 is KC = 2.7 x 10^-17 at some temperature T.
A system initially has [N2] = 0.0800 M and [O2] = 0.0500 M. There is no NO initially present in
the system. Find the value for [NO] that will be present at equilibrium.
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