Question

Calculate the equilibrium constant Kp for this reaction, given the following information (at 299 K ): 2NO(g)+Br2(g)⇌2NOBr(g)Kc=2.2 2NO(g)⇌N2(g)+O2(g)Kc=2.3×1030

Answer 1

Answer:

$2NO_{(g)}+Br_2_{(g)}\rightleftharpoons2NOBr_{(g)}$ , Kp = 0.08967

$2NO_{(g)}\rightleftharpoons N_2_{(g)}+O_2_{(g)}$ , Kp = 2.3×10³⁰

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

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:

$2NO_{(g)}+Br_2_{(g)}\rightleftharpoons2NOBr_{(g)}$

Given: Kc = 2.2

Temperature = 299 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2+1) = -1

Thus, Kp is:

$K_p= 2.2\times (0.082057\times 299)^{-1}$

Kp = 0.08967

For the second equilibrium reaction:

$2NO_{(g)}\rightleftharpoons N_2_{(g)}+O_2_{(g)}$

Given: Kc = 2.3×10³⁰

Temperature = 299 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2) = 0

Thus, Kp is:

$K_p= 2.2\times (0.082057\times 299)^{0}$

Kp =  2.3×10³⁰

Answer 2

The equilibrium constant is proportionate of the products partial pressure to that of the reactants. The equilibrium constant (Kp) for the first reaction is 0.08967 and for a second, $2.3 \times 10^{30}$.

What is the equilibrium constant Kp of reaction?

The equilibrium constant can be estimated as the proportion between the pressure or the concentration of yields and reactants of the chemical reaction.

The relation between Kp and Kc is:

$\rm K_{p} = K_{c} \times (RT)^{\Delta n}$

Here, Kp = pressure equilibrium constant, Kc = molar equilibrium constant, R = gas constant and T = temperature in Kelvins

For the first reaction:

Given,

• Temperature = 299 K
• Kc = 2.2
• R = $0.082057 \rm \; L \;atm\;mol^{-1}K^{-1}$
• Δn = -1

Kp can be calculated as:

$\begin{aligned}\rm K_{p} &= 2.2 \times (0.082057 \times 299)^{-1}\\\\&= 0.0896\end{aligned}$

For the second reaction:

Given,

• Kc = $2.3 \times 10^{30}$
• Temperature = 299 K
• R = $0.082057 \rm \; L \;atm\;mol^{-1}K^{-1}$
• Δn = 0

Kp can be calculated as:

$\begin{aligned}\rm K_{p} &= 2.3 \times 10^{30} \times (0.082057 \times 299)^{0}\\\\&= 2.3 \times 10^{30}\end{aligned}$

Therefore, the equilibrium constant for the first reaction is 0.08967 and for a second, $2.3 \times 10^{30}$.

Learn more about the equilibrium constant here:

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