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Consider the following equilibrium for which kc=420 at 24 c

N2(g) + O2(g) + Br2(g) <---> 2NOBr(g)
What is the value of Kp for this reaction?

User Rjdown
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1 Answer

5 votes

Answer:

Step-by-step explanation:

To find the value of Kp for the given reaction,

relationship between Kp and Kc for gaseous reactions, which is:

Kp = Kc(RT)^(Δn)

where R is the gas constant, T is the temperature in Kelvin, and Δn is the difference between the sum of the stoichiometric coefficients of the products and the sum of the stoichiometric coefficients of the reactants.

In this case, the equation is:

N2(g) + O2(g) + Br2(g) ⇌ 2NOBr(g)

The stoichiometric coefficients for the products (NOBr) add up to 2, and the stoichiometric coefficients for the reactants (N2, O2, Br2) add up to 4. Therefore, Δn = 2 - 4 = -2.

Substituting the values into the equation for Kp, we get:

Kp = Kc(RT)^(Δn)

= 420 (0.0821 L·atm/mol·K)^(−2) (24+273 K)^(−2)

Kp= 0.695 atm^2

Therefore, the value of Kp for the given reaction is 0.695 atm^2.

User Sallie
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