Problem 45 - Chemistry: Principles and Reactions

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 \(P_{\mathrm{C}_{2} \mathrm{~N}_{2}}=P_{\mathrm{H}_{2}}=0.32 \mathrm{~atm}, P_{\mathrm{HCN}}=0.45 \mathrm{~atm} ?\)

Step-by-Step Solution

Answer

Answer: The equilibrium partial pressures are as follows: - Hydrogen cyanide (HCN): \(0.1112 ~\mathrm{atm}\) - Cyanogen (\(\mathrm{C}_{2} \mathrm{N}_{2}\)): \(0.4894 ~\mathrm{atm}\) - Hydrogen (\(\mathrm{H}_{2}\)): \(0.4894 ~\mathrm{atm}\)

Step 1: Write the equilibrium expression for the given reaction

From the balanced chemical equation, $$2 \mathrm{HCN}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{~N}_{2}(g) + \mathrm{H}_{2}(g)$$ The equilibrium expression can be written as: $$K = \frac{\left[\mathrm{C}_{2} \mathrm{~N}_{2}\right] \left[\mathrm{H}_{2}\right]}{\left[\mathrm{HCN}\right]^2}$$

Step 2: Define the change in pressure in terms of a variable

Let the change in pressure during the reaction be x. So, at equilibrium, the change in partial pressures of the gases will be: - Hydrogen cyanide: \(0.45 - 2x~\mathrm{atm}\) - Cyanogen: \(0.32 + x~\mathrm{atm}\) - Hydrogen: \(0.32 + x~\mathrm{atm}\)

Step 3: Substitute the equilibrium pressures into the equilibrium expression and solve for x

Substituting the equilibrium pressures into the equilibrium expression, we can write: $$0.17 = \frac{(0.32 + x)(0.32 + x)}{(0.45 - 2x)^2}$$ Solving this quadratic equation will give us the value of x.

Step 4: Solve for x

After solving the quadratic equation, we find that there are two possible solutions for x: 1. \(x = 0.1694 ~\mathrm{atm}\) 2. \(x = -0.4794 ~\mathrm{atm}\) However, because a negative change in pressure does not make physical sense in this context, we can discard the second solution.

Step 5: Calculate the equilibrium partial pressures

Using the value of \(x = 0.1694 ~\mathrm{atm}\), we can find the equilibrium partial pressures of the gases: - Hydrogen cyanide: \(P_\mathrm{HCN} = 0.45 - 2(0.1694) = 0.45 - 0.3388 = 0.1112 ~\mathrm{atm}\) - Cyanogen: \(P_\mathrm{C_{2}N_{2}} = 0.32 + 0.1694 = 0.4894 ~\mathrm{atm}\) - Hydrogen: \(P_\mathrm{H_{2}} = 0.32 + 0.1694 = 0.4894 ~\mathrm{atm}\) So at equilibrium, \(P_{\mathrm{C}_{2} \mathrm{~N}_{2}} = P_{\mathrm{H}_{2}} = 0.4894 \mathrm{~atm}\) and \(P_{\mathrm{HCN}} = 0.1112 \mathrm{~atm}\).

Key Concepts

Equilibrium Expression

One of the fundamental concepts in chemical equilibrium is the equilibrium expression, which relates the concentrations of reactants and products at equilibrium. It is derived from a balanced chemical equation and follows the law of mass action.

For the given reaction where hydrogen cyanide decomposes into cyanogen and hydrogen gas, the equilibrium expression is written based on the coefficients of the balanced equation. To illustrate, the expression \(K = \frac{[\mathrm{C}_{2} \mathrm{~N}_{2}] [\mathrm{H}_{2}]}{[\mathrm{HCN}]^2}\) is a ratio of the products' concentrations (or partial pressures in the case of gases) to the reactants' concentrations, each raised to the power of their respective coefficients in the balanced chemical equation.

When writing an equilibrium expression, remember the following:

Partial Pressure

In gas-phase reactions, the term partial pressure is used instead of concentration. Partial pressure is the pressure each gas in a mixture exerts if it were alone in the container and is proportional to its mole fraction. When gases are involved in the equilibrium, the equilibrium expression is adjusted to include the partial pressures.

In our problem, we started with initial partial pressures for cyanogen (\(\mathrm{C}_{2} \mathrm{N}_{2}\)), hydrogen (\(\mathrm{H}_{2}\)), and hydrogen cyanide (\(\mathrm{HCN}\)). The changes in partial pressures caused by the reaction are represented by a variable \(x\), and they directly impact the equilibrium state. As the reaction progresses towards equilibrium, the partial pressures shift accordingly. Moreover, discovering how these pressures change and calculating the final equilibrium partial pressures are instrumental in understanding how systems respond to change.

Equilibrium Constant

The equilibrium constant, represented by \(K\), is a numerical value that signifies the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their stoichiometric coefficients. This constant is specific to a particular reaction at a fixed temperature.

In the exercise, we found that the equilibrium constant for the decomposition of hydrogen cyanide is \(0.17\). This constant tells us the extent to which a reaction moves forward; a small equilibrium constant indicates that the reactants are favored at equilibrium, while a large value suggests the products are favored.

Understanding the precise meaning of \(K\) is essential for making predictions about chemical reactions. For example, from the value of \(K\) and the initial partial pressures, we can predict the direction in which the reaction will proceed to reach equilibrium. By applying the mathematical relationship inherent in \(K\), we can calculate the changes needed to reach this state and thus determine the final partial pressures of all gases involved.