Answer:Equilibrium constant, Kp, for the reaction is 0.00227
Explanation:
We are given:
The chemical reaction for the decomposition follows the equation:
[tex]2NOBr(g)\rightleftharpoons 2NO(g)+Br_2(g)[/tex]
At t = 0 0.260 0 0
At [tex]t=t_{eq}[/tex] 0.260-2x 2x x
The expression for [tex]K_p[/tex] for the given reaction follows:
[tex]K_p=\frac{(p_{NO})^2\times p_{Br_2}}{(p_{NOBr})^2}[/tex]
We are given:
[tex]2x=\frac{22}{100}\times 0.260=0.0572[/tex]
[tex]x=0.0286[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{(2x)^2\times x}{(0.260-2x)^2}[/tex]
[tex]K_p=\frac{(0.0572)^2\times 0.0286}{(0.260-0.0572)^2}[/tex]
[tex]K_p=0.00227[/tex]
Thus equilibrium constant, Kp, for the reaction is 0.00227
We have that for the Question "What is the equilibrium constant, Kp, for the reaction [tex]2NOBr(g) ↔ 2NO(g) + Br_2[/tex]"
Answer:
From the question we are told
0.260 atm of NOBr is sealed in a flask and allowed to reach equilibrium, 22% of the NOBr decomposes
2NOBr ↔ 2NO(g) + Br_2
0.26 0 0
-2X +2X X
[tex]= \frac{22*0.26}{100} = 2X\\\\= 0.0572 =2X[/tex]
At equilibrium,
[tex]NOBr = 0.26-0.0572 = 0.2028atm\\\\NO = 0.0572atm\\\\Br_2 = \frac{0.0572}{2} =0.0286[/tex]
Therefore,
[tex]K_p = (0.0572)^2 * \frac{0.0286}{(0.203)^2}\\\\K_p = 2.27*10^{-3}[/tex]
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