Problem 40 - Chemistry The Science in Context

Question

How is the value of the equilibrium constant \(K_{\mathrm{p}}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) $$ related to the value of \(K_{\mathrm{p}}\) for this reaction at the same temperature? $$ \mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{N}_{2}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{NO}(g) $$

Step-by-Step Solution

Answer

Answer: The relationship between the equilibrium constants is \(K_{\mathrm{p1}} = K_{\mathrm{p2}}^{2}\).

Step 1: Determine the first reaction's \(K_{\mathrm{p}}\) expression

For the first reaction, represented by: $$2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g),$$ the expression for the equilibrium constant \(K_{\mathrm{p}}\) can be written as: $$K_{\mathrm{p1}}=\frac{[\mathrm{H}_{2}]^{2}[\mathrm{NO}]^{2}}{[\mathrm{H}_{2}\mathrm{O}]^{2}[\mathrm{N}_{2}]}.$$

Step 2: Determine the second reaction's \(K_{\mathrm{p}}\) expression

For the second reaction, represented by: $$\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{N}_{2}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{NO}(g),$$ the expression for the equilibrium constant \(K_{\mathrm{p}}\) can be written as: $$K_{\mathrm{p2}}=\frac{[\mathrm{H}_{2}][\mathrm{NO}]}{[\mathrm{H}_{2}\mathrm{O}][\mathrm{N}_{2}]^{\frac{1}{2}}}.$$

Step 3: Find the relation between \(K_{\mathrm{p1}}\) and \(K_{\mathrm{p2}}\)

Given the expressions for \(K_{\mathrm{p1}}\) and \(K_{\mathrm{p2}}\), we see that both expressions have the same numerator, but have a different exponent for the denominator. To find a relation between \(K_{\mathrm{p1}}\) and \(K_{\mathrm{p2}}\), we must set the second reaction as the square root of the first reaction. We can rewrite the second reaction by squaring the equation as: $$2\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2\mathrm{H}_{2}(g)+2\mathrm{NO}(g),$$ which is the same as the first reaction. Now, when we square the \(K_{\mathrm{p2}}\) expression, we have: $$K_{\mathrm{p2}}^{2} = \frac{[\mathrm{H}_{2}]^{2}[\mathrm{NO}]^{2}}{[\mathrm{H}_{2}\mathrm{O}]^{2}[\mathrm{N}_{2}]}.$$

Step 4: Compare the expressions for \(K_{\mathrm{p1}}\) and \(K_{\mathrm{p2}}^2\)

We see that the expression for \(K_{\mathrm{p2}}^{2}\) is equivalent to the expression for \(K_{\mathrm{p1}}\). Therefore, the relation between \(K_{\mathrm{p1}}\) and \(K_{\mathrm{p2}}\) is: $$K_{\mathrm{p1}} = K_{\mathrm{p2}}^{2}.$$

Key Concepts

Reaction Kinetics

Understanding reaction kinetics involves studying the speed or rate of chemical reactions and the factors that affect this rate. In simple terms, reaction kinetics tells us how fast or slow a reaction occurs based on things like temperature, concentration of reactants, and the presence of a catalyst.

Several factors can affect reaction rates:

• Concentration of Reactants: Increasing the concentration generally increases the rate of reaction because more particles are available to collide and react.
• Temperature: Higher temperatures lead to more energetic particles, which collide more frequently, thus speeding up the reaction.
• Catalysts: A catalyst speeds up a reaction without being consumed in the process. It provides an alternative pathway with a lower activation energy.
• Surface Area: The larger the surface area of a reactant, the faster the reaction goes, as more particles are exposed for collisions.

The study of kinetics is crucial for understanding how to control and optimize different chemical processes, whether they're happening in a laboratory or in industrial settings.

Chemical Equilibrium

Chemical equilibrium refers to a state in a reversible reaction where the rates of the forward and backward reactions are equal, meaning the concentration of reactants and products remains constant over time. It's like a perfect balance in a see-saw where both sides are even.

Key aspects of chemical equilibrium include:

• Dynamic Process: At equilibrium, the reaction doesn't stop. Both forward and backward reactions keep occurring at equal rates.
• Equilibrium Constant, \(K\): For reactions at equilibrium, the ratio of concentrations of products to reactants, each raised to the power of their coefficients, is constant at a given temperature.
• Le Chatelier’s Principle: If a system at equilibrium is disturbed, it will adjust itself to counteract the disturbance and restore a new equilibrium state.
• Position of Equilibrium: This term describes whether the concentration of products or reactants is higher when equilibrium is achieved.

Understanding equilibrium gives insight into how different conditions can shift the balance of a reaction to favor the formation of products or reactants, which is vital for controlling industrial and natural chemical processes.

Gas-Phase Reactions

Gas-phase reactions are chemical reactions where the reactants and products are in the gaseous state. These reactions are notable for their fast rates because gases mix well and collide more frequently than solids or liquids.

Important features of gas-phase reactions include:

• Ideal Gas Law: Many gas-phase reactions are described by the ideal gas law, \(PV = nRT\), which connects pressure (\(P\)), volume (\(V\)), amount of substance (\(n\)), and temperature (\(T\)).
• Pressure and Volume Effects: Changes in pressure and volume can significantly impact the reaction kinetics and equilibrium. Increasing pressure can push the equilibrium towards the side with fewer gas molecules.
• Homogeneous Reactions: As all reactants and products are gases, many gas-phase reactions are homogeneous, occurring in a single phase, leading to uniform reaction conditions.

Due to their nature, gas-phase reactions are essential in both natural processes such as atmospheric reactions and industrial settings, including the production of ammonia through the Haber process. Understanding these reactions helps in managing pollution and manufacturing chemicals efficiently.