Respuesta :
Answer:
1) 0.1437 is the equilibrium constant,[tex]K_p[/tex] of this reaction.
2) [tex]9.867\times 10^{-6}[/tex]is the equilibrium constant,[tex]K_c[/tex] of this reaction.
Explanation:
1) [tex]CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)[/tex]
At equilbrium :
Partial pressure of the CO gas = [tex]p_1=0.840 atm[/tex][
Partial pressure of the [tex]Cl_2[/tex] gas = [tex]p_2=1.16 atm[/tex]
Partial pressure of the [tex]COCl_2[/tex] gas = [tex]p_3=0.140 atm[/tex]
Expression of[tex]K_p[/tex] is given by :
[tex]K_p=\frac{p_3}{p_1\times p_2}[/tex]
[tex]=\frac{0.140 atm}{0.840 atm\times 1.16 atm}=0.1437 [/tex]
0.1437 is the equilibrium constant,[tex]K_p[/tex] of this reaction.
2) [tex]H_2(g)+I_2(g)\rightleftharpoons 2HI(g)[/tex]
Initial:
4.00 M 2.55 M
At equilibrium ;
(4.00-0.0100) M (2.55-0.0100) M (0.0100) M
(3.9900) M (2.5400) M (0.0100) M
Expression of[tex]K_c[/tex] is given by :
[tex]K_c=\frac{[HI]^2}{[H_2][I_2]}[/tex]
[tex]=\frac{(0.0100 M)^2}{3.9900 M\times 2.5400 M}=9.867\times 10^{-6}[/tex]
[tex]9.867\times 10^{-6}[/tex]is the equilibrium constant,[tex]K_c[/tex] of this reaction.
Phosgene is the organic chemical compound with the formula COCl₂. It is a colorless gas; in low concentrations, its odor resembles that of freshly cut hay or grass.
The formula we will use is as follows:-
[tex]k_p = \frac{p_3}{P_2*P_1}[/tex].
The partial pressure of the[tex]CO,\ CL_2,\ and \ COCL_2[/tex] is given in the question. after putting the value into the formula.
[tex]k_p = \frac{0.140}{0.840*1.16}[/tex] , After solving it the value of Kp is 0.1437.
Hence, 0.1437 is the equilibrium constant.
In the second problem, the equilibrium constant is:-
[tex]k_c=\frac{{HI}^2}{H_2*I_2}[/tex], after putting the value,
[tex]k_c=\frac{{0.01}^2}{3.9900*2.5400}[/tex]
The equilibrium constant is:- [tex]9.867 *10^{-6}[/tex]
For more information, refer to the link:-
https://brainly.com/question/14530482
