Question

For a first order homogeneous gaseous reaction, \(A \longrightarrow 2 B+C\) then initial pressure was \(P_{i}\) while total pressure after time ' \(t\) ' was \(P_{t}\). The right expression for the rate constants \(k\) in terms of \(P_{i}, P_{t}\) and \(t\) is : (a) \(k=\frac{2.303}{t} \log \left(\frac{2 P_{i}}{3 P_{i}-P_{t}}\right)\) (b) \(k=\frac{2.303}{t} \log \left(\frac{2 P_{i}}{2 P_{t}-P_{i}}\right)\) (c) \(k=\frac{2.303}{t} \log \left(\frac{P_{i}}{P_{i}-P_{t}}\right)\) (d) none of these

Short answer

The correct expression for the rate constant \(k\) is option (a): \(k=\frac{2.303}{t} \log \left(\frac{2 P_{i}}{3 P_{i}-P_{t}}\right)\).

Step-by-step solution

- Write the Stoichiometry

The stoichiometry of the given reaction is: \(A \longrightarrow 2B + C\). This indicates that one mole of A produces two moles of B and one mole of C.

- Relate Change in Moles to Change in Pressure

As the reaction progresses, pressure changes due to the change in the number of moles. Initially, the pressure is \(P_i\) due to substance A. At time 't', substance A will have partially converted into B and C, increasing the number of moles and thus the pressure becomes \(P_t\).

- Calculate Moles at Time 't'

Let the moles of A reacted be x. Initially, there are 0 moles of B and C. According to the stoichiometry, x moles of A will produce 2x moles of B and x moles of C. Since the initial moles of A are represented by the initial pressure \(P_i\), the moles reacted can be related to the pressures by \(x = P_i - \frac{P_t}{3}\).

- Establish the Rate Expression

The rate of the reaction can be expressed in terms of the change in concentration (or pressure for gases) over time. Thus, rate = \(\frac{kx}{t}\).

- Solve for k

Rearrange the rate expression to find the rate constant \(k\): \(k = \frac{rate \times t}{x}\). We can substitute the value of x from Step 3 and the rate expression from Step 4 into this equation.

- Apply the Initial Conditions and Solve for k

By inserting the expressions for rate and x, and using the logarithmic relation between concentration and pressure changes for first order reactions, the correct expression for k is \(k = \frac{2.303}{t} \log \left(\frac{2 P_{i}}{3 P_{i} - P_{t}}\right)\), which corresponds to option (a).

Key concepts

Reaction Stoichiometry

Understanding the stoichiometry of a reaction is crucial in determining how reactants transform into products. In the given first order reaction, \(A \longrightarrow 2B + C\), the stoichiometry tells us that one mole of substance A produces two moles of B and one mole of C. This proportional relationship is foundational for predicting how much product will form when a certain amount of reactant is used.

Furthermore, reaction stoichiometry helps in balancing chemical equations, which is vital for calculations involving the conservation of mass. It ensures that for every mole of A consumed, the expected moles of products are formed, thereby allowing us to relate changes in pressure with moles of reactants and products.

Pressure Change in Reactions

In gaseous systems, pressure is a direct indication of the number of moles present, given a constant temperature and volume. During a reaction, as reactants convert to products, the mole count may change, and this alters the pressure. For a homogeneous gaseous reaction like the provided example, an increase in the number of moles will generally lead to an increase in pressure.

In simple terms, if more gas particles are produced, they will exert a higher pressure. Hence, pressure changes in the reaction can provide valuable insights into the progress of the reaction over time. This association between pressure and the number of moles is governed by the Ideal Gas Law, which represents one of the applications in chemical kinetics and reaction monitoring.

Calculating Rate Constant

The rate constant, \(k\), is a parameter that quantifies the speed of a chemical reaction. Calculating the rate constant is a key step in characterizing a reaction’s kinetics. For first order reactions, the rate is proportional to the concentration of one reactant. When dealing with gases, pressure can serve as a proxy for concentration.

By rearranging the rate expression to isolate the rate constant, and substituting in values relating the pressures to moles of reactants consumed and products formed, we obtain an expression for \(k\). It’s important to conceptualize this as connecting how swiftly a reactant is consumed to the rise in pressure of the system, as demonstrated in the step-by-step solution. This calculation of \(k\) provides critical information for predicting how long a reaction will take to reach a certain point, or how reactant concentrations will change over time.

Relationship Between Pressure and Moles

The relationship between pressure and moles in a gaseous system can be described by the Ideal Gas Law, \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for moles, \(R\) for the universal gas constant, and \(T\) for temperature. When temperature and volume are held constant, the pressure of a gas is directly proportional to the number of moles of gas present.

In the context of our reaction, the initial pressure \(P_i\) is due to the moles of reactant A present. As the reaction proceeds, A is converted into products B and C, increasing the total number of moles and therefore the pressure, captured at any time by \(P_t\). Understanding this relationship allows us to link pressure measurements back to mole calculations, enabling us to monitor the reaction's progress and calculate the rate at which reactants are consumed.