Problem 64

Question

Pure nitrosyl chloride (NOCl) gas was heated to $240^{\circ} \mathrm{C}$ in a $1.00-\mathrm{L}$ container. At equilibrium the total pressure was 1.00 atm and the $\mathrm{NOCl}$ pressure was 0.64 atm. $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate the partial pressures of $\mathrm{NO}$ and $\mathrm{Cl}_{2}$ in the system. (b) Calculate the equilibrium constant $K_{P}$.

Step-by-Step Solution

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Answer

The partial pressures of $\mathrm{NO}$ and $\mathrm{Cl}_{2}$ at equilibrium are 0.36 atm and 0.18 atm respectively. The equilibrium constant $K_{P}$ for the reaction at $\mathrm{240^{\circ} C}$ is 0.15625.

1Step 1: Analyze the reaction and given data

Firstly, write down the stoichiometrically balanced chemical reaction, as it is given: $2 \mathrm{~NOCl}(g) \rightleftharpoons 2 \mathrm{~NO}(g)+\mathrm{Cl}_{2}(g)$. Analyze the given data. It is known that the total pressure at equilibrium is 1.00 atm, and the $\mathrm{NOCl}$ pressure is 0.64 atm.

2Step 2: Calculate partial pressures of $\mathrm{NO}$ and $\mathrm{Cl}_{2}$

In the equilibrium reaction, 2 moles of $\mathrm{NOCl}$ decompose to form 2 moles of $\mathrm{NO}$ and 1 mole of $\mathrm{Cl}_{2}$. This implies that for every 1 atm of pressure drop in $\mathrm{NOCl}$, there should be an increase of 0.5 atm in $\mathrm{Cl}_{2}$ and 1 atm in $\mathrm{NO}$, keeping the total pressure at equilibrium constant. Mathematically, $P_{\mathrm{NO}} = P_{\mathrm{NOCl, initial}} - P_{\mathrm{NOCl, final}} = 1.00~\mathrm{atm} - 0.64~\mathrm{atm} = 0.36~\mathrm{atm}$ and $P_{\mathrm{Cl}_{2}} = P_{\mathrm{NO}}/2 = 0.36~\mathrm{atm}/2 = 0.18~\mathrm{atm}$. So, the partial pressures of $\mathrm{NO}$ and $\mathrm{Cl}_{2}$ are 0.36 atm and 0.18 atm respectively.

3Step 3: Calculate the equilibrium constant $K_{P}$

The equilibrium constant $K_{P}$ 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 a power equal to its stoichiometric coefficient. Based on the balanced reaction equation, we have: $K_{P} = \frac{(P_{\mathrm{NO}})^{2} \cdot P_{\mathrm{Cl}_{2}}}{(P_{\mathrm{NOCl}})^{2}} = \frac{(0.36)^{2} \cdot 0.18}{(0.64)^{2}} = 0.15625$. The value of $K_{P}$ is 0.15625.

Key Concepts

Partial Pressure in Chemical EquilibriaUnderstanding the Equilibrium Constant $ K_P $The Role of Stoichiometry

Partial Pressure in Chemical Equilibria

Partial pressure is a concept that helps us understand the contribution of each gas in a mixture to the total pressure. It's simply the pressure a gas would exert if it alone occupied the entire volume. In this problem, we are dealing with a gas mixture at equilibrium. The total pressure is given as 1.00 atm, and $NOCl$ is specifically at 0.64 atm. To find the partial pressures of other gases, think about the reaction: when $NOCl$ decomposes, it forms $NO$ and $Cl_2$.
Let's break this down into two steps:

Firstly, the total change in pressure from $NOCl$ decomposition is linked to how much $NO$ and $Cl_2$ form. Each mole of $NOCl$ gives rise to one mole of $NO$ and half a mole to $Cl_2$.
The pressure drop from 1.00 atm to 0.64 atm tells us how much $NO$ and $Cl_2$ are generated. So, $0.36 \, \mathrm{atm}$ worth of $NO$ and $0.18 \, \mathrm{atm}$ worth of $Cl_2$ are formed, based on stoichiometry.

Understanding partial pressures helps balance the equation of the gases involved, maintaining the total equilibrium pressure.

Understanding the Equilibrium Constant $ K_P $

The equilibrium constant, $ K_P $, is a core concept in understanding chemical equilibria, particularly for reactions involving gases. It's essentially a number that describes the ratio of concentrations (or pressures) of products and reactants at equilibrium.

In our problem, we calculate $ K_P $ using partial pressures. The equation from the balanced reaction is the key:

The formula is \[ K_{P} = \frac{(P_{\mathrm{NO}})^{2} \cdot P_{\mathrm{Cl}_{2}}}{(P_{\mathrm{NOCl}})^{2}} \]
The powers are based on the coefficients from the balanced reaction: square terms for $NO$ and linear for $Cl_2$, reflecting their stoichiometric quantities.
Substituting the partial pressures: \[ K_{P} = \frac{(0.36)^{2} \cdot 0.18}{(0.64)^{2}} = 0.15625 \]

This value reflects the position of equilibrium: a small $ K_P $ suggests reactants are favored, while a large $ K_P $ would indicate products.

The Role of Stoichiometry

Stoichiometry is the essence of quantitative relationships in chemical reactions. It tells us how much of each substance is consumed and produced. This concept helps us assess the changes in the chemical system quantitatively. In this exercise, stoichiometry reveals critical insights:

The balanced equation $2 NOCl \rightleftharpoons 2 NO + Cl_2$ shows a 2:2:1 stoichiometric relationship. This means for every 2 moles of $NOCl$ decomposed, 2 moles of $NO$ and 1 mole of $Cl_2$ are produced.
Based on this stoichiometry, for every pressure unit drop of $NOCl$, there's a corresponding creation of pressure due to $NO$ and $Cl_2$. Hence, our change in moles also reflects changes in pressure due to the constant volume condition.

Stoichiometry is the bridge that connects the qualitative description of a reaction to the quantitative changes in conditions, helping us calculate partial pressures and equilibrium constants accurately.

Problem 63

Problem 65