Question

At a certain temperature, \(K_{\mathrm{c}}=4.3 \times 10^{5}\) for the reaction $$ \mathrm{HCO}_{2} \mathrm{H}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ If \(0.200 \mathrm{~mol}\) of \(\mathrm{HCO}_{2} \mathrm{H}\) is placed in a \(1.00 \mathrm{~L}\) vessel, what will be the concentrations of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) when the system reaches equilibrium? (Hint: Where does the position of equilibrium lie when \(K\) is very large?)

Short answer

The concentrations of CO and \(H_2O\) at equilibrium are both approximately 0.200 M

Step-by-step solution

Setting up the ICE Table

To determine the equilibrium concentrations, use an ICE (Initial, Change, Equilibrium) table. Initially, the concentration of \(HCO_2H\) is 0.200 M, and both products are 0 M. As the reaction proceeds to the right (due to a large \(K_c\)), \(x\) moles of \(HCO_2H\) will react to form \(x\) moles of CO and \(H_2O\). The equilibrium concentrations will be 0.200 - \(x\) for \(HCO_2H\), and \(x\) for CO and \(H_2O\).

Writing the expression for Kc

The equilibrium constant expression for the reaction is \[K_{\mathrm{c}} = \frac{[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]}{[\mathrm{HCO}_{2}\mathrm{H}]}\]. Plug in the equilibrium concentrations in terms of \(x\) and the given \(K_c\) value to solve for \(x\).

Solving for x considering the magnitude of Kc

The large value of \(K_c\) (\(4.3 \times 10^{5}\)) implies that the reaction strongly favors the product side. Therefore, we can assume that all the \(HCO_2H\) is converted to products and that \(x\) is approximately 0.200. This assumption is valid unless the calculated \(x\) is more than 5% of the initial concentration, in which case we would need to use the quadratic formula.

Checking the assumption and calculating equilibrium concentrations

Verify the assumption by calculating \(x\) using the equilibrium expression. If \(x\) is much smaller than 0.200, the assumption holds. Calculate the concentrations of CO and \(H_2O\) as both will be \(x\) i.e., approximately 0.200 M.

Key concepts

Equilibrium Constant

The equilibrium constant, represented by the symbol \(K_c\) in the context of concentration, is a quantitative measure of the position of equilibrium for a reversible chemical reaction. This constant is crucial as it indicates the relative amounts of reactants and products present at equilibrium. For example, a large value of \(K_c\), as given in the exercise with \(4.3 \times 10^{5}\), tells us that at equilibrium, the concentration of products is much higher than the concentration of reactants. This implies the reaction favors the formation of products, which will be of primary interest when predicting the outcome of the reaction.

Key to understanding equilibrium constant is that it is temperature-dependent and specific to a particular reaction at a constant temperature. This means that \(K_c\) will change if the temperature changes but remain constant if other conditions like pressure or concentration are altered, provided temperature is constant. For a balanced reaction such as \(\mathrm{HCO}_{2}\mathrm{H}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_{2}\mathrm{O}(g)\), the equilibrium constant expression in terms of concentration is given by \[ K_{\mathrm{c}} = \frac{[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]}{[\mathrm{HCO}_{2}\mathrm{H}]} \.\] Understanding how to formulate and use this expression is foundational for predicting the direction and extent of chemical reactions.

ICE Table

An ICE table stands for Initial, Change, and Equilibrium, which is a systematic method to organize the concentrations of reactants and products during a chemical reaction. This table becomes particularly helpful when dealing with equilibrium problems. Starting with initial concentrations, one can determine the changes that occur as the system moves towards equilibrium, and finally the equilibrium concentrations of all species involved.

To construct an ICE table, list the reactants and products down one side and then enter the initial concentrations, the changes that occur during reaction, and the equilibrium concentrations in separate columns. The change typically involves unknown values (often denoted by \(x\)) that reflect the stoichiometry of the balanced chemical equation. As illustrated in the exercise, with an initial concentration of \(0.200\,\mathrm{mol/L}\) for \(\mathrm{HCO}_{2}\mathrm{H}\) and zero for the products CO and \(\mathrm{H}_{2}\mathrm{O}\), one calculates the changes in concentrations when the reaction reaches equilibrium, and those changes are depicted as plus or minus \(x\), which then allows us to solve for \(x\) and thus find the equilibrium concentrations.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a measure similar to the equilibrium constant, but for any point in time during a reaction, not just at equilibrium. It is calculated using the same formula as the \(K_c\) but with the concentrations or partial pressures of the reactants and products at that specific moment, rather than at equilibrium.

The value of \(Q\) can indicate the direction in which a reaction will need to shift to reach equilibrium. If \(Q < K_c\), the reaction will proceed in the forward direction to produce more products. Conversely, if \(Q > K_c\), the reaction will proceed in the reverse direction to produce more reactants. When \(Q = K_c\), the reaction is at equilibrium, and no net change will occur. In the exercise provided, the initial reaction quotient is zero since no products are present initially, and the reaction will shift towards the product side because \(Q < K_c\), supporting the concept of where the position of equilibrium lies when \(K\) is very large.