Question

The equilibrium constant \(K_{\mathrm{c}}\) for the following reaction is 0.65 at \(395^{\circ} \mathrm{C}\). $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ (a) What is the value of \(K_{P}\) for this reaction? (b) What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?\) (c) What is \(K_{\mathrm{c}}\) for \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g) ?\) (d) What are the values of \(K_{P}\) for the reactions described in (b) and (c)?

Short answer

The answers: (a) \(K_{P}\) for the first reaction can be calculated by substituting given values into the formula \(K_p = K_c(RT)^{\Delta n}\). (b) The equilibrium constant \(K_{\mathrm{c}}\) for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. So, \(K_{\mathrm{c}} = 1/0.65\). (c) For the reaction \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\), \(\mathrm{K}_{\mathrm{c}} = \sqrt{0.65}\) . (d) For reaction in (b), use the \(K_{_{\mathrm{c}}}\) calculated in (b) in the formula \(K_p = K_c(RT)^{\Delta n}\) to get \(K_{P}\). Similarly, for reaction in (c), use the \(K_{_{\mathrm{c}}}\) calculated in (c) in the formula to get \(K_{P}\).

Step-by-step solution

Calculate the value of \(K_{P}\)

We know that \(K_p = K_c(RT)^{\Delta n}\), where \( \Delta n = n_{products} - n_{reactants}\) , \(R\) is the gas constant (\(0.0821 \, L \, atm \, mol^{-1} \, K^{-1}\)) and \(T\) is the temperature in Kelvin. Here, \( \Delta n = 2 - (1 + 3) = -2 \), \( K_c = 0.65 \) and \(T = 395^\circ C = 395 + 273.15 = 668.15 K \). Substitute these values into the formula to get \( K_p \).

Calculate the value of \(K_{\mathrm{c}}\) for the reverse reaction

The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. So, for the reverse reaction \(2\mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\), the equilibrium constant \( K_{\mathrm{c}} = 1/ K_{\mathrm{c}}_{forward} = 1/0.65 \).

Calculate the value of \(K_{\mathrm{c}}\) for the reaction with different stoichiometry

If a reaction is multiplied by a factor, the equilibrium constant is raised to that factor. Here, for reaction \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\) which is half the original reaction, \(\mathrm{K}_{\mathrm{c}} = (K_{\mathrm{c}}_{original})^{\sfrac{1}{2}}= (0.65)^{\sfrac{1}{2}} = \sqrt{0.65} \).

Calculate the value of \(K_{p}\) for the reactions in (b) and (c)

In case (b), apply the steps from Step 1 with the new equilibrium constant obtained in Step 2. In case (c), apply the steps from Step 1 with the new equilibrium constant obtained in Step 3 and with newly calculated \(\Delta n\) for this reaction.

Key concepts

Kc and Kp relationship

Understanding the relationship between the equilibrium constants, specifically, the relationship of the constants for reactions involving gases—denoted as K_c and K_p—is critical in balancing chemical reactions and predicting their behavior under changing conditions. K_c refers to the equilibrium constant expressed in terms of concentration, while K_p is the equilibrium constant expressed in terms of partial pressures.

To connect K_c with K_p, we use the equation K_p = K_c(RT)^Δn, where Δn is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants), R is the gas constant, and T is the temperature in Kelvin. The ability to switch between K_c and K_p allows chemists to predict how reactions will shift when conditions like pressure and volume are altered.

Reaction Quotient

The reaction quotient, Q, is like a 'snapshot' of a reaction at any point in time and can help determine which direction a reaction will proceed to reach equilibrium. Q is calculated in the same way as the equilibrium constant K_c but isn't exclusive to the state of equilibrium. Instead, Q uses the current concentrations of the reactants and products.

Comparing Q to the equilibrium constant tells us about the system's state: if Q < K_c, the reaction will proceed forward to create more products; if Q > K_c, the reaction will go in reverse to create more reactants. If Q = K_c, the reaction is at equilibrium, and the concentrations of reactants and products will remain stable unless external changes are made.

Le Chatelier's Principle

Le Chatelier's Principle is a valuable rule of thumb for predicting the effect of a change in conditions on chemical equilibria. It states that if an external change is imposed on a system at equilibrium, the system will adjust itself to counteract the change and re-establish equilibrium.

For instance, an increase in pressure will favor the direction that produces fewer moles of gas. Similarly, a temperature change will affect the equilibrium position depending on whether the reaction is exothermic or endothermic. This principle is essential for optimizing yields in chemical manufacturing and understanding natural processes.

Chemical Equilibrium

Chemical equilibrium represents a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products over time. It is important to note that even though the macroscopic properties remain constant, the molecular processes are dynamic—reactants continuously form products and products revert to reactants at the same rate.

At equilibrium, the equilibrium constant K provides a ratio of the concentrations of products to reactants raised to their stoichiometric coefficients. Knowing the value of K can help chemists understand the extent to which a reaction will favor the formation of products or reactants, crucial for predicting the outcomes of reactions in various industries and natural phenomena.