Question

Hydrogen cyanide, a highly toxic gas, can decompose to cyanogen and hydrogen gases, $$ 2 \mathrm{HCN}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{~N}_{2}(g)+\mathrm{H}_{2}(g) $$ At a certain temperature, \(K\) for this decomposition is 0.17 . What are the partial pressures of all gases at equilibrium if initially the partial pressures are \(P_{\mathrm{C}_{2} \mathrm{~N}_{2}}=P_{\mathrm{H}_{2}}=0.32 \mathrm{~atm}\) \(P_{\mathrm{HCN}}=0.45 \mathrm{~atm} ?\)

Short answer

A: The partial pressures at equilibrium are: - \(P_{\mathrm{HCN}} = 0.15 \ \mathrm{atm}\) - \(P_{\mathrm{C}_{2}\mathrm{N}_{2}} = 0.47 \ \mathrm{atm}\) - \(P_{\mathrm{H}_{2}} = 0.47 \ \mathrm{atm}\)

Step-by-step solution

Write the ICE table

An ICE table is a tool used to keep track of the changes of concentrations/pressures in chemical reactions involving gases at equilibrium. The table is organized as follows: | | HCN | C2N2 | H2 | |----------|-----------|---------|----------| | Initial | 0.45 atm | 0.32 atm | 0.32 atm | | Change | -2x | +x | +x | | Equilibrium | 0.45-2x | 0.32+x | 0.32+x | The "x" represents the change in the pressures of the gases, which is unknown right now.

Write the equilibrium expression

The equilibrium expression is based on the equilibrium constant \(K\). In this exercise, K is given as 0.17. Write the equilibrium expression as follows: $$K = \frac{[C_{2}N_{2}][H_{2}]}{[HCN]^2}$$

Substitute the equilibrium values from the ICE table

Now, substitute the values from the ICE table into the equilibrium expression, using partial pressures for the gases: $$0.17 = \frac{(0.32+x)(0.32+x)}{(0.45-2x)^2}$$

Solve the quadratic equation for x

Solving the quadratic equation can be done using a variety of methods, depending on the student's level and preference. In this case, we can use an iterative method or a calculator to find that: $$x = 0.15$$

Calculate the final partial pressures

Now that we have the value of x, we can plug it back into the equilibrium concentrations in the ICE table to find the final pressures: \(P_{\mathrm{HCN}} = 0.45 - 2(0.15) = 0.15\ \mathrm{atm}\) \(P_{\mathrm{C}_{2}\mathrm{N}_{2}} = 0.32 + 0.15 = 0.47\ \mathrm{atm}\) \(P_{\mathrm{H}_{2}} = 0.32 + 0.15 = 0.47\ \mathrm{atm}\) The partial pressures at equilibrium are as follows: - \(P_{\mathrm{HCN}} = 0.15 \ \mathrm{atm}\) - \(P_{\mathrm{C}_{2}\mathrm{N}_{2}} = 0.47 \ \mathrm{atm}\) - \(P_{\mathrm{H}_{2}} = 0.47 \ \mathrm{atm}\)

Key concepts

ICE Table

Understanding the changes in reactants and products during a chemical reaction that has reached equilibrium is a breeze with the ICE table. It stands for Initial, Change, and Equilibrium – these are the stages that reveal how the reactant and product concentrations evolve in a reaction. Let's say we start with known initial pressures of the gases involved. As the reaction progresses, these pressures shift, leading to a change, denoted as 'x' for unknown amounts that are lost or gained.

At equilibrium, the pressures stabilize, and we can express them in terms of this 'x'. For any given reaction, we can set up an ICE table to systematically keep track of these shifts. This proves incredibly useful in solving for equilibrium concentrations or partial pressures, as seen in the hydrogen cyanide decomposition problem. By populating this table with the initial pressures and the changes each component undergoes, as well as the resulting equilibrium pressures, we have a clear map of the reaction's journey towards equilibrium.

Equilibrium Constant

The equilibrium constant, represented by the symbol K, is a value that gives us crucial insight into the behavior of a reversible chemical reaction at equilibrium. It's determined by the ratio of the concentrations (or partial pressures) of the products to reactants at equilibrium, each raised to the power of their respective stoichiometric coefficients.

In our problem, we're dealing with the decomposition of hydrogen cyanide, and the equilibrium constant, K, is given as 0.17. Knowing this allows us to predict the direction of the reaction and to calculate unknown equilibrium concentrations. It essentially acts as the 'goal' for how product and reactant pressures will balance out when the reaction is at rest in a closed system.

Partial Pressures

In gas-phase reactions, the 'partial pressure' is the pressure that each gas in a mixture would exert if it alone occupied the entire volume of the mixture. It's an essential concept in chemical equilibrium because the equilibrium constant for reactions involving gases can be expressed in terms of partial pressures.

Initially, each gas has a defined partial pressure, but as the reaction reaches equilibrium, these pressures adjust. For our hydrogen cyanide reaction, once we solve for 'x' using the equilibrium constant and the ICE table, we can determine the equilibrium partial pressures of hydrogen cyanide, cyanogen, and hydrogen gas. This provides a snapshot of the gas mixture at equilibrium and helps us understand the extent to which the reaction has proceeded.

Equilibrium Expression

The equilibrium expression is the mathematical embodiment of a reaction's equilibrium state. It's based directly on the balanced chemical equation and incorporates the equilibrium constant. For reactions involving gases, like the decomposition of hydrogen cyanide, the equilibrium expression is crafted using partial pressures.

We write it as a quotient, where the numerator contains the products of the reaction, each raised to the power of its stoichiometric coefficient, and the denominator contains the reactants in the same manner. Plugging in the equilibrium pressures from the ICE table into this expression allows us to solve for the unknown changes during the reaction. For students, understanding how to translate a chemical equation into an equilibrium expression is key to unraveling equilibrium problems.