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In a chemical equilibrium rate constant of forward reaction is \(7.5 \times 10^{-4}\) and the equilibrium constant is 1.5. The rate constant of backward reaction is: (a) \(2.5 \times 10^{4}\) (b) \(5 \times 10^{-4}\) (c) \(2.5 \times 10^{-4}\) (d) \(5 \times 10^{4}\)

Short Answer

Expert verified
The rate constant of the backward reaction is \( 5 \times 10^{-4} \) (option b).

Step by step solution

01

Understanding the Relationship

For a chemical reaction in equilibrium, the equilibrium constant, denoted as \( K \), is given by the ratio of the rate constant of the forward reaction (\( k_f \)) to the rate constant of the backward reaction (\( k_b \)). The relationship is:\[ K = \frac{k_f}{k_b} \]
02

Substituting Given Values

We are given that the equilibrium constant \( K = 1.5 \) and the forward reaction rate constant \( k_f = 7.5 \times 10^{-4} \). We need to substitute these values into the equilibrium equation:\[ 1.5 = \frac{7.5 \times 10^{-4}}{k_b} \]
03

Rearranging the Equation

To find \( k_b \), rearrange the equation to solve for it. This can be done by multiplying both sides of the equation by \( k_b \):\[ 1.5 \times k_b = 7.5 \times 10^{-4} \]
04

Solving for the Backward Rate Constant

Now divide both sides by 1.5 to solve for \( k_b \): \[ k_b = \frac{7.5 \times 10^{-4}}{1.5} \]Calculate the result of this division.
05

Finding the Numerical Solution

Perform the division:\[ k_b = \frac{7.5}{1.5} \times 10^{-4} = 5 \times 10^{-4} \]Therefore, the rate constant of the backward reaction is \( 5 \times 10^{-4} \).

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Key Concepts

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

Rate Constant
The rate constant is a fundamental concept in chemical kinetics. It indicates the speed of a chemical reaction and is denoted by the symbol \(k\). For a given reaction, the rate constant differs between forward and backward reactions.
When a reaction moves forward, it transforms reactants into products, and this is defined by the forward rate constant \(k_f\). Conversely, when the reaction proceeds in the reverse direction, converting products back into reactants, it's characterized by the backward rate constant \(k_b\).
  • These constants are units of rate reaction, typically in terms of concentration over time (e.g., \(\text{mol/L/s}\)).
  • An important aspect of rate constants is their dependency on temperature. As temperature increases, so does the rate constant, leading to faster reactions.
  • Knowing the rate constants helps to predict the speed of a reaction at a given temperature and provides insights into the reaction mechanism itself.
Equilibrium Constant
The equilibrium constant, symbolized as \(K\), is a vital indicator of a system's behavior in a state of chemical equilibrium. It represents the ratio of the concentrations of products to reactants, each raised to the power of their coefficients as found in the balanced chemical equation.
  • For a generic reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is given by \[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
  • An equilibrium constant value greater than 1 indicates that, at equilibrium, the products are favored.
  • Conversely, if \( K \) is less than 1, the reactants are favored when equilibrium is reached.
  • Equilibrium constants do not have units. Instead, they reflect the ratio and relative strength of forward and backward rate constants, specifically \( K = \frac{k_f}{k_b} \).
Determining the equilibrium constant provides comprehensive understanding not only of where the reaction conditions stand but also infers the position of equilibrium. This is crucial in predicting the extent of reactions under specific circumstances.
Forward and Backward Reactions
Forward and backward reactions are two integral facets of reversible chemical reactions. In any given equilibrium system, the forward reaction converts reactants into products, while the backward reaction turns products back into reactants.
In a system in equilibrium, the rates of forward and backward reactions are precisely balanced, resulting in constant concentrations of products and reactants. This underscores the concept of dynamic equilibrium, where reactions proceed in both directions but no net change is observed.
  • The forward reaction is often driven by the reactants' inherent chemical energy and the conditions such as concentration and temperature.
  • Backward reaction rate is influenced by the energy and concentration of the products.
  • The concept of forward and backward reactions provides insight into reaction dynamics and how equilibrium can shift with changes in conditions like temperature and pressure (principle of Le Chatelier).
The interplay of these reactions establishes the equilibrium point, a central concept in understanding chemical equilibria and predicting how a system's composition might change in response to external changes.

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Most popular questions from this chapter

Phosphorous pentachloride dissociates as follows, in a closed reaction vessel. \(\mathrm{PCI}_{5}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\). If total pressure at equilibrium of the reaction mixture is \(\mathrm{P}\) and degree of dissociation of \(\mathrm{PC} 1_{5}\) is \(x\), the partial pressure of \(\mathrm{PCl}_{3}\) will be: (a) \(\left(\frac{x}{(x+1)}\right) \mathrm{P}\) (b) \(\left(\frac{2 x}{(x-1)}\right) \mathrm{P}\) (c) \(\left(\frac{x}{(x-1)}\right) \mathrm{P}\) (d) \(\left(\frac{x}{(1-x)}\right) \mathrm{P}\)

An amount of solid \(\mathrm{NH}_{4}\) HS in placed in a flask already containing ammonia gas at a certain temperature and \(0.50\) atm pressure. Ammonium hydrogen sulphide decomposes to yield \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure is the flask rises to \(0.84 \mathrm{~atm}\), the equilibrium constant for \(\mathrm{NH}_{4} \mathrm{HS}\) decomposition at this temperature is: (a) \(0.30\) (b) \(0.18\) (c) \(0.17\) (d) \(0.11\)

For the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\), the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) are 4 and \(8 \mathrm{~atm}\) respectively. The value of \(K_{p}\) for this reaction is: (a) \(14 \mathrm{~atm}\) (b) \(16 \mathrm{~atm}\) (c) \(18 \mathrm{~atm}\) (d) \(12 \mathrm{~atm}\)

In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? = equilibrium constant): (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.001\) (b) \(\mathrm{M} \rightleftharpoons \mathrm{N} ; \mathrm{K}=10\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=0.005\) (d) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=0.01\)

The vapour density of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at a certain temperature is 30\. What is the percentage dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at this temperature: (a) \(53.3\) (b) \(106.6\) (c) \(26.7\) (d) None of these

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