The equilibrium constant for the reaction: 2NO(g) + Br2(g)<----> 2NOBr(g) is Kc = 1.3×10^-2 at…
The following solution is suggested to handle the subject “The equilibrium constant for the reaction: 2NO(g) + Br2(g)<----> 2NOBr(g) is Kc = 1.3×10^-2 at…“. Let’s keep an eye on the content below!
Question “The equilibrium constant for the reaction: 2NO(g) + Br2(g)<----> 2NOBr(g) is Kc = 1.3×10^-2 at…”
The equilibrium constant for the reaction: 2NO(g) + Br2(g)
<—-> 2NOBr(g) is Kc = 1.3×10^-2 at 1,000 K
a.) At this temperature, does the equilibrium favor the product
or reactants?
b.) Calculate Kc for 2NOBr <—-> 2NO + Br2
c.) Calculate Kc for NOBr <—-> NO + 1/2Br2
Answer
This concept is based upon equilibrium constant ( -2NOBr(g)-is-Kc-=-1.3×10^-2-at.png?x-oss-process=image/resize,w_560/format,webp” rel=”nofollow” target=”_blank”>
Below is the general reaction to equilibrium.
-2NOBr(g)-is-Kc-=-1.3×10^-2-at.png?x-oss-process=image/resize,w_560/format,webp” rel=”nofollow” target=”_blank”>
(1)
Below is the equilibrium reaction.
-2NOBr(g)-is-Kc-=-1.3×10^-2-at.png?x-oss-process=image/resize,w_560/format,webp” rel=”nofollow” target=”_blank”>
-2NOBr(g)-is-Kc-=-1.3×10^-2-at.png?x-oss-process=image/resize,w_560/format,webp” rel=”nofollow” target=”_blank”>
#media_tag_18$ in equation (2).
-2NOBr(g)-is-Kc-=-1.3×10^-2-at.png?x-oss-process=image/resize,w_560/format,webp” rel=”nofollow” target=”_blank”>
-2NOBr(g)-is-Kc-=-1.3×10^-2-at.png?x-oss-process=image/resize,w_560/format,webp” rel=”nofollow” target=”_blank”>
#media_tag_24$ in equation (II).
Above is the solution for “The equilibrium constant for the reaction: 2NO(g) + Br2(g)<----> 2NOBr(g) is Kc = 1.3×10^-2 at…“. We hope that you find a good answer and gain the knowledge about this topic of science.