Problem 35 - Chemistry: The Central Science in SI Units, Global Edition

Question

A mixture of \(0.140 \mathrm{~mol}\) of \(\mathrm{NO}, 0.060 \mathrm{~mol}\) of \(\mathrm{H}_{2},\) and 0.260 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a \(2.0-\mathrm{L}\) vessel at \(330 \mathrm{~K}\). Assume that the following equilibrium is established: $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ At equilibrium \(\left[\mathrm{H}_{2}\right]=0.010 \mathrm{M}\). (a) Calculate the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2},\) and \(\mathrm{H}_{2} \mathrm{O} .\) (b) Calculate \(K_{c}\).

Step-by-Step Solution

Answer

The equilibrium concentrations are \(NO:~0.030~M\), \(N_2:~0.020~M\), and \(H_2O:~0.170~M\). The equilibrium constant, \(K_c \approx 637\).

Step 1: Determine Change in Concentrations

: Let's define the change in concentrations for the species involved in the reaction as they reach equilibrium: - \(\Delta [NO] = -2x\) - \(\Delta [H_2] = -x\) - \(\Delta [N_2] = +x\) - \(\Delta [H_2O] = +2x\) The problem gives us that the \(equilibrium~concentration~of~H_2 = 0.010~M\) \[Initial~concentration~of~H_2 + \Delta [H_2] = Equilibrium~concentration~of~H_2\]

Step 2: Calculate the Change (x)

: Now, solve for x: \(0.030 - x = 0.010\) \((x = 0.020~M)\)

Step 3: Calculate Equilibrium Concentrations

: Using the value of x, we can now calculate the equilibrium concentrations of the other species: \[Equilibrium~concentration~of~NO = (0.070 - 2(0.020)) = 0.030~M\] \[Equilibrium~concentration~of~N_2 = (0 + 0.020) = 0.020~M\] \[Equilibrium~concentration~of~H_2O = (0.130 + 2(0.020)) = 0.170~M\]

Step 4: Calculate Kc

: Now, we can calculate the equilibrium constant, Kc, using these equilibrium concentrations: \(K_c = \frac{[N_2][H_2O]^2}{[NO]^2[H_2]^2}\) \(K_c = \frac{(0.020)(0.170)^2}{(0.030)^2(0.010)^2}\) \(K_c \approx 637\) - The equilibrium concentrations are \(NO:~0.030~M\), \(N_2:~0.020~M\), and \(H_2O:~0.170~M\). - The equilibrium constant, \(K_c \approx 637\).

Key Concepts

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), is a measure that tells us the extent to which a chemical reaction proceeds to form products at equilibrium. It is derived from the equilibrium concentrations of products and reactants in a balanced reaction. The reaction given in the exercise is:\[2 \mathrm{NO}(g) + 2 \mathrm{H}_2(g) \rightleftharpoons \mathrm{N}_2(g) + 2 \mathrm{H}_2 \mathrm{O}(g)\]For this reaction, \(K_c\) is calculated using the formula:\[ K_c = \frac{[\mathrm{N}_2][\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{NO}]^2[\mathrm{H}_2]^2} \]The units are determined such that they depend on the change in the number of moles of gas. If we have equilibrium concentrations of the species:- \([\mathrm{N}_2] = 0.020\, \mathrm{M}\)- \([\mathrm{H}_2\mathrm{O}] = 0.170\, \mathrm{M}\)- \([\mathrm{NO}] = 0.030\, \mathrm{M}\)- \([\mathrm{H}_2] = 0.010\, \mathrm{M}\)We plug these values into the formula to find \(K_c\):\[K_c = \frac{(0.020)(0.170)^2}{(0.030)^2(0.010)^2} \approx 637\]This indicates that at equilibrium, the concentration of products is much higher than that of reactants, suggesting a reaction favoring the formation of products.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept to understand how a system at equilibrium responds to disturbances. It states that if an external change is applied to a system at equilibrium, the system adjusts itself to partially counteract that change.For example, in our reaction, if the concentration of \(\mathrm{NO}\) is increased, the system will try to reduce this increase by consuming more \(\mathrm{NO}\) to form more \(\mathrm{N}_2\) and \(\mathrm{H}_2\mathrm{O}\). Thus, the reaction shifts to the right.This principle helps predict how:

• Changes in concentration of reactants or products
• Changes in pressure (for gaseous reactions)
• Changes in temperature influence equilibrium

Understanding Le Chatelier's Principle allows chemists to control reactions to maximize product yield by applying specific changes to the system.

Reaction Quotient

The Reaction Quotient, \(Q\), gives insight into the direction in which a reaction is progressing. While \(K_c\) is only applicable at equilibrium, \(Q\) can be calculated at any point to predict the shift needed to reach equilibrium.Similar to \(K_c\), \(Q\) for the reaction is given by:\[ Q = \frac{[\mathrm{N}_2][\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{NO}]^2[\mathrm{H}_2]^2} \]**Using \(Q\) to Assess Progress:**- **If** \(Q > K_c\): More products are present, the reaction will shift to the left to form more reactants.- **If** \(Q < K_c\): More reactants are present, the reaction will shift to the right to form more products.- **If** \(Q = K_c\): The system is at equilibrium and no shift is needed.Suppose at some point, the concentrations were calculated to yield a \(Q\) less than 637, this would indicate the reaction is not at equilibrium and would proceed towards more products until \(Q\) equals \(K_c\). Understanding \(Q\) helps chemists to monitor and control reactions effectively.