Question

Consider the equilibrium system described by the chemical reaction below. For this reaction, Kp = 10^9 at 298 K. If the equilibrium partial pressures of Brz and NOBr at 0.0159 atm and 0.0768 atm, respectively, determine the partial pressure of NO at equilibrium. 2NO + Br2 ⇌ 2NOBr

Answer 1

The equilibrium partial pressure of NO at 298 K is approximately 2.43 x 10^(-3) atm.

Step-by-step explanation:

Chemical equilibrium is governed by the equilibrium constant (Kp), which is the ratio of the product of the partial pressures of the products to the product of the partial pressures of the reactants, each raised to the power of their stoichiometric coefficients. For the given reaction:

`\[2NO + Br_2 \rightleftharpoons 2NOBr\]`

The equilibrium constant (Kp) is given as 10^9, which can be expressed as:

`\[Kp = \frac{{[NOBr]^2}}{{[NO]^2 [Br_2]}}\]`

Given the equilibrium partial pressures of Br_2 and NOBr as 0.0159 atm and 0.0768 atm, respectively, we can use these values to determine the equilibrium partial pressure of NO. Let \(x\) be the equilibrium partial pressure of NO; thus, the equilibrium concentrations become
`\([NO] = [NO]_0 - 2x\), \([Br_2] = [Br_2]_0 - x\), and \([NOBr] = [NOBr]_0 + 2x\)`.

Substituting these values into the equilibrium constant expression and solving the quadratic equation, we find x, the equilibrium partial pressure of NO, to be approximately 2.43 x 10^-3 atm. Therefore, at 298 K, the partial pressure of NO at equilibrium is 2.43 x 10^-3 atm. This calculation ensures the system is in a state where the rate of the forward and reverse reactions are equal, achieving chemical equilibrium.