Respuesta :
Answer:
[tex]2NO_{(g)}+Br_2_{(g)}\rightleftharpoons2NOBr_{(g)} [/tex]
[tex]K_p= 0.074[/tex]
[tex]2NO_{(g)}\rightleftharpoons N_2_{(g)}+O_2_{(g)}[/tex]
[tex]K_p= 2.3\times 10^{30}[/tex]
Explanation:
The relation between Kp and Kc is given below:
[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
[tex]2NO_{(g)}+Br_2_{(g)}\rightleftharpoons2NOBr_{(g)} [/tex]
Given: Kc = 1.8
Temperature = 296 K
R = 0.082057 L atm.mol⁻¹K⁻¹
Δn = (2)-(2+1) = -1
Thus, Kp is:
[tex]K_p= 1.8\times (0.082057\times 296)^{-1}[/tex]
[tex]K_p= 0.074[/tex]
For the second equilibrium reaction:
[tex]2NO_{(g)}\rightleftharpoons N_2_{(g)}+O_2_{(g)}[/tex]
Given: Kc = [tex]2.3\times 10^{30}[/tex]
Temperature = 296 K
R = 0.082057 L atm.mol⁻¹K⁻¹
Δn = (2)-(2) = 0
Thus, Kp is:
[tex]K_p= 2.3\times 10^{30}\times (0.082057\times 296)^{0}[/tex]
[tex]K_p= 2.3\times 10^{30}[/tex]
