The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{P}_{4}(g) \rightleftharpoons 2 \mathrm{P}_{2}(g) $$ is \(1.4\) at \(400^{\circ} \mathrm{C}\). Suppose that 3 moles of \(\mathrm{P}_{4}(g)\) and 2 moles of \(\mathrm{P}_{2}(g)\) are mixed in 2 litre container at \(400^{\circ} \mathrm{C}\). What is the value of reaction quotient \(\left(Q_{c}\right)\) ? (a) \(\frac{3}{2}\) (b) \(\frac{2}{3}\) (c) 1 (d) None of these

Short Answer

Expert verified
\(Q_c = \frac{2}{3}\) and the correct answer is (b).

Step by step solution

01

Understanding the Reaction Quotient (\(Q_c\))

The reaction quotient, \(Q_c\), is calculated using the expression that is similar to the equilibrium constant (\(K_c\)) but with the initial concentrations of the reactants and products, not at equilibrium. For the reaction \(\text{P}_4(g) \rightleftharpoons 2 \text{P}_2(g)\), the expression for \(Q_c\) is \[Q_c = \frac{[\text{P}_2]^2}{[\text{P}_4]}\] where the brackets denote the concentration of the substance.
02

Calculating the Initial Concentrations

Given that 3 moles of \(\text{P}_4\) and 2 moles of \(\text{P}_2\) are mixed in a 2 litre container, their initial concentrations are: \[\text{Concentration of } \text{P}_4 = \frac{3 \text{ moles}}{2 \text{ liters}} = 1.5 \text{ M}\] \[\text{Concentration of } \text{P}_2 = \frac{2 \text{ moles}}{2 \text{ liters}} = 1 \text{ M}\]
03

Inserting the Concentrations into the \(Q_c\) Expression

Insert the initial concentrations of \(\text{P}_4\) and \(\text{P}_2\) into the \(Q_c\) expression to obtain \(Q_c\): \[Q_c = \frac{(1 \text{ M})^2}{1.5 \text{ M}} = \frac{1}{1.5} = \frac{2}{3}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
In the study of chemical reactions, chemical equilibrium is a crucial concept. It refers to the state in a reaction where the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of the reactants and products over time. This dynamic equilibrium can be approached from either direction of the reaction.

Imagine two teams playing tug-of-war with equal strength. They pull back and forth, but neither side wins because their power balances out; similarly, in a chemical reaction at equilibrium, molecules continue to react, but at a rate that sustains an unvarying ratio of product to reactant concentrations. Essential factors that can influence equilibria include concentration, temperature, and pressure.
Equilibrium Constant
The equilibrium constant, symbolized as \( K_c \) responsible for quantitative analysis. It is a ratio derived from the concentrations of the products raised to the power of their stoichiometric coefficients to the concentrations of the reactants raised to the power of their stoichiometric coefficients, at equilibrium.

In the given reaction \( \text{P}_4(g) \rightleftharpoons 2 \text{P}_2(g) \) the equilibrium constant \( K_c \) is 1.4 at \( 400^\text{o}C \) which tells us that at this temperature, the concentration of the products, \( \text{P}_2 \) squared, over the concentration of the reactant, \( \text{P}_4 \) will always yield a value of 1.4 when the system is at equilibrium. This constant is intrinsic to a reaction at a given temperature and does not change unless the temperature shifts.

Understanding \( K_c \) is essential as it predicts the direction of the reaction; a higher \( K_c \) indicates a larger concentration of products at equilibrium, thus favoring the forward reaction, whereas a lower \( K_c \) favors the reverse reaction.
Initial Concentrations
The initial concentrations of substances in a chemical reaction are the concentrations before any reaction has taken place. They are especially significant when predicting how the reaction will progress towards equilibrium.

In our example, we began with 3 moles of \( \text{P}_4 \) and 2 moles of \( \text{P}_2 \) in a 2 liter container, resulting in initial concentrations of 1.5 M for \( \text{P}_4 \) and 1 M for \( \text{P}_2 \) respectively. These initial concentrations are plugged into the reaction quotient \( Q_c \), which is analogous to \( K_c \), but for non-equilibrium conditions.

The calculation of \( Q_c \) with these starting concentrations allows us to predict how the system will change to achieve equilibrium. If \( Q_c < K_c \), the system will shift towards the products to reach equilibrium. Conversely, if \( Q_c > K_c \) implies a shift towards the reactants. It's a forecasting tool used to determine the initial direction of the shift towards equilibrium.

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Most popular questions from this chapter

A catalyst increases the rate of a reaction by: (a) increasing the activation energy of the reaction (b) increasing the value of rate constants \(\left(k_{f}\right.\) and \(k_{b}\) ) (c) increasing the enthalpy change of the reaction (d) decreasing the enthalpy change of the reaction

\(9.2\) grams of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is taken in a closed one litre vessel and heated till the following equilibrium is reached \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) At equilibrium, \(50 \% \mathrm{~N}_{2} \mathrm{O}_{4}(g)\) is dissociated. What is the equilibrium constant (in mol litre \(^{-1}\) ) (molecular weight of \(\mathrm{N}_{2} \mathrm{O}_{4}=92\) ) (a) \(0.1\) (b) \(0.4\) (c) \(0.2\) (d) 2

The value of \(\Delta G^{\circ}\) for a reaction \(^{\circ}\) in aqueous phase having \(K_{c}=1\), would be : (a) \(-R T\) (b) \(-1\) (c) 0 (d) \(+R T\)

Two moles of \(\mathrm{NH}_{3}\) when put into a previously evacuated vessel (one litre), partially dissociated into \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\). If at equilibrium one mole of \(\mathrm{NH}_{3}\) is present, the equilibrium constant is : (a) \(3 / 4 \mathrm{~mol}^{2}\) litre \(^{-2}\) (b) \(27 / 64 \mathrm{~mol}^{2}\) litre \(^{-2}\) (c) \(27 / 32 \mathrm{~mol}^{2}\) litre \(^{-2}\) (d) \(27 / 16 \mathrm{~mol}^{2}\) litre \(^{-2}\)

At \(273 \mathrm{~K}\) and 1 atm, 10 litre of \(\mathrm{N}_{2} \mathrm{O}_{4}\) decomposes to \(\mathrm{NO}_{2}\) according to equation $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ What is degree of dissociation \((\alpha)\) when the original volume is \(25 \%\) less than that of existing volume? (a) \(0.25\) (b) \(0.33\) (c) \(0.66\) (d) \(0.5\)

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