Question

Consider the reaction $\mathrm{A}(g)+2\mathrm{B}(g)\rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$ in a 1.0-L rigid flask. Answer the following questions for each situation $(\mathrm{a}-\mathrm{d})$ : i. Estimate a range (as small as possible) for the requested substance. For example, [A] could be between $95M$ and $100M$ ii. Explain how you decided on the limits for the estimated range. iii. Indicate what other information would enable you to narrow your estimated range. iv. Compare the estimated concentrations for a through d, and explain any differences. a. If at equilibrium $[\mathrm{A}]=1M$, and then $1\mathrm{mol}\mathrm{C}$ is added, estimate the value for $[\mathrm{A}]$ once equilibrium is reestablished. b. If at equilibrium $[\mathrm{B}]=1M$, and then $1\mathrm{mol}\mathrm{C}$ is added, estimate the value for [B] once equilibrium is reestablished. c. If at equilibrium $[\mathrm{C}]=1M$, and then $1\mathrm{mol}\mathrm{C}$ is added, estimate the value for $[\mathrm{C}]$ once equilibrium is reestablished. d. If at equilibrium $[\mathrm{D}]=1M$, and then $1\mathrm{mol}\mathrm{C}$ is added, estimate the value for [D] once equilibrium is reestablished.

Short answer

In summary, for the given reaction, the estimated concentration ranges after adding 1 mol of C are: a. [A] should be between 1 M and some value greater than 1 M. b. [B] should be between 1 M and some value less than 1 M. c. [C] should be between 1 M and 2 M. d. [D] should be between 1 M and some value less than 1 M. These estimations are based on Le Chatelier's Principle and calculated reaction quotients (Q'). To further narrow down these ranges, information on initial concentrations and equilibrium constants (K) would be needed.

Step-by-step solution

(Situation a: calculation)

In this case, at equilibrium, [A] = 1 M and we are adding 1 mol of C. When we add 1 mol of C, the reaction will shift to the left to re-establish equilibrium (Le Chatelier's Principle). To find the new equilibrium value for [A], we can set up an ICE table: Initial: [A] = 1 M, [B] = x, [C] = 1+x, [D] = y (we don't know the initial concentrations of B and D) Change: [A] = +z, [B] = +2z, [C] = -z, [D] = -z Equilibrium: [A] = 1+z, [B] = x+2z, [C] = x, [D] = y-z The reaction quotient (Q) will be defined as: $$Q=\frac{[C][D]}{[A][B{]}^2}$$ After adding 1 mol of C, we get the new reaction quotient Q': $$Q^{\prime }=\frac{(1+x)(y-z)}{(1+z)(x+2z{)}^2}$$ Since we want the reaction to shift to the left, and considering that Q' > Q, the new equilibrium concentration [A] will be higher than 1 M. Thus, the estimated range for [A] should be between 1 M and some value greater than 1 M.

(Situation a: explanation and other information)

We decided on the estimated range for [A] based on Le Chatelier's Principle and by calculating the reaction quotient Q and Q'. To narrow down our estimated range, we would need more information about the equilibrium constant (K) and the initial concentrations of B and D.

(Situation b: calculation)

In this case, at equilibrium, [B] = 1 M and we are adding 1 mol of C. The reaction will shift to the left, as in situation a. We can set up another ICE table: Initial: [A] = x, [B] = 1, [C] = 1+x, [D] = y Change: [A] = +z, [B] = -2z, [C] = -z, [D] = -z Equilibrium: [A] = x+z, [B] = 1-2z, [C] = x, [D] = y-z By calculating Q' again, we get: $$Q^{\prime }=\frac{(1+x)(y-z)}{(x+z)(1-2z{)}^2}$$ In this case, since we want the reaction to shift to the left, the new equilibrium concentration [B] should be between 1 M and some value less than 1 M.

(Situation b: explanation and other information)

The estimated range for [B] is based on the calculation of Q' and the same understanding of Le Chatelier's Principle. To narrow down the estimated range, we would need the initial concentrations of A and D and the equilibrium constant K.

(Situation c: calculation)

In this case, at equilibrium, [C] = 1 M, and we are adding 1 mol of C making the concentration of C to be [C]=2M initially. The reaction will shift to the left. We can set up an ICE table: Initial: [A] = x+z, [B] = y, [C] = 2, [D] = z Change: [A] = +z, [B] = +2z, [C] = -z, [D] = -z Equilibrium: [A] = x+2z, [B] = y+2z, [C] = 2-z, [D] = z By calculating Q' again, we get: $$Q^{\prime }=\frac{(2-z)(z)}{(x+2z)(y+2z{)}^2}$$ In this case, C will need to decrease for equilibrium to be reestablished (shift to the left). So, the new equilibrium concentration of [C] should be between 1 M and 2 M.

(Situation c: explanation and other information)

The estimated range for [C] is based on the calculation of Q' and an understanding of Le Chatelier's Principle. To narrow down the estimated range, we would need the initial concentrations of A and B and the equilibrium constant K.

(Situation d: calculation)

In this case, at equilibrium, [D] = 1 M, and we are adding 1 mol of C. The reaction will shift to the left. We can set up an ICE table: Initial: [A] = x+z, [B] = y+2z, [C] = 1+x, [D] = 1 Change: [A] = +z, [B] = +2z, [C] = -z, [D] = -z Equilibrium: [A] = x+2z, [B] = y+4z, [C] = 1, [D] = 1-z By calculating Q' again, we get: $$Q^{\prime }=\frac{1}{(x+2z)(y+4z{)}^2}$$ In this case, since we want the reaction to shift to the left, the new equilibrium concentration [D] should be between 1 M and some value less than 1 M.

(Situation d: explanation and other information)

The estimated range for [D] is based on the calculation of Q' and an understanding of Le Chatelier's Principle. To narrow down the estimated range, we would need the initial concentrations of A and B and the equilibrium constant K.

(Comparison of the situations)

Comparing the estimated concentrations for situations a through d, we can see that when a substance in the reaction is increased in concentration, the reaction will shift in a direction to counteract the change (Le Chatelier's Principle). Specifically, we can notice that: - In situations a and b, when [A] or [B] is initially at 1 M, adding 1 mol of C prompts a shift to the left, increasing the concentrations of A and decreasing the concentrations of B. - In situation c, when [C] is initially at 1 M, adding 1 mol of C causes the reaction to shift to the left, decreasing the concentration of C while increasing the concentrations of A and B. - In situation d, when [D] is initially at 1 M, adding 1 mol of C causes the reaction to shift to the left, decreasing the concentrations of D while increasing the concentrations of A and B.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This happens because the forward and reverse reactions occur at equal rates. In essence, the reaction has reached a point where there is no overall change in the system.
Even though concentrations don't change, the microscopic processes are still ongoing. For example, in the reaction of $$A(g)+2B(g)\rightleftharpoons C(g)+D(g),$$ even when equilibrium is reached, molecules of A and B continue to form C and D, while C and D revert back to A and B at the same rate. Understanding chemical equilibrium is essential for predicting how changes like pressure, temperature, or concentration affect a reaction. Le Chatelier's Principle explains how systems in equilibrium adjust to counteract changes. So, if you add a substance to a system at equilibrium, the reaction will shift in a direction that reduces the effect of the added substance.

ICE Table Methodology

The ICE table is a handy tool for visualizing changes in concentration during a chemical reaction approaching equilibrium. ICE stands for Initial, Change, and Equilibrium. Here's how it works:

• Initial: Write the initial concentrations of all species involved in the reaction. This is the starting point before any shifts in the reaction occur.
• Change: Determine how the concentrations will change as the system shifts to toward equilibrium. Usually, changes in concentration are denoted with variables like ±x or ±z.
• Equilibrium: Calculate the equilibrium concentrations by applying the changes to the initial concentrations.

For example, if you know the initial concentrations at equilibrium and add 1 mol of C, you can use the ICE table to calculate how the reaction shifts and find the new equilibrium concentrations. In each scenario (a-d), the ICE table assists in detailing these shifts step by step, making it easier to predict how each component will adjust and what the new equilibrium state will be.

Reaction Quotient

The reaction quotient, Q, helps determine the direction in which a reaction at non-equilibrium concentrations will proceed to reach equilibrium. It is calculated the same way as the equilibrium constant, K, but with current concentrations instead of equilibrium concentrations.
For the reaction $A(g)+2B(g)\rightleftharpoons C(g)+D(g)$, the reaction quotient is given by:$$Q=\frac{[C][D]}{[A][B{]}^2}$$By comparing Q to K, you can predict the shift:

• If $Q<K$, the reaction proceeds forward to produce more products and reach equilibrium.
• If $Q>K$, the reaction will shift to the left, favoring the formation of reactants to achieve equilibrium.
• If $Q=K$, the system is already at equilibrium, and no shift occurs.

In the scenarios provided, Q' helps determine how the addition of a substance like C affects the shift in the equilibrium position back towards equilibrium.

Equilibrium Constant

The equilibrium constant (K) is a crucial factor in understanding chemical equilibrium. It quantifies the ratio of product concentrations to reactant concentrations at equilibrium, raised to the power of their respective coefficients in the balanced equation. For our reaction:$$K=\frac{[C][D]}{[A][B{]}^2}$$The magnitude of K indicates the position of equilibrium:

• If $K\gg 1$, the products are favored at equilibrium.
• If $K\ll 1$, the reactants are favored.

Knowing K helps to predict how factors like a change in concentration will affect the equilibrium. For practical problems like the ones presented, the precise value of K allows a more accurate calculation of the exact shift in concentrations. Unfortunately, without a known equilibrium constant, only estimated ranges can be given for the new equilibrium states after changes like the addition of a mole of C.