Q.For a reaction, aA + bB -> products, -d[A]/dt is equal to

In chemical kinetics, the rate of a reaction can be expressed in terms of the rate of disappearance of reactants or the rate of formation of products. However, these individual rates are not always equal to each other because of the stoichiometry of the reaction. For example, if 3 moles of A disappear for every 2 moles of B formed, then A is disappearing faster than B is forming (in terms of moles).

To define a single, unambiguous "rate of reaction" for the entire process, we normalize the rate of change of concentration of each species by its stoichiometric coefficient. This ensures that the overall reaction rate is independent of which reactant or product we choose to monitor.

Let's apply this concept to the given reaction.

1. Write the general rate expression for the given reaction. The reaction is $3A \rightarrow 2B$. Using the general formula, the rate of reaction can be expressed in terms of the disappearance of A and the formation of B:

Here, $\frac{d[A]}{dt}$ represents the instantaneous rate of change of concentration of A, and $\frac{d[B]}{dt}$ represents the instantaneous rate of change of concentration of B. The negative sign for A indicates it is a reactant and its concentration is decreasing. The positive sign for B indicates it is a product and its concentration is increasing.

2. Equate the relevant parts of the rate expression and solve for the desired term. …

Understanding Reaction Order

The order of a reaction tells us how the rate depends on reactant concentrations. It's a purely experimental quantity—you cannot deduce it from the stoichiometric equation alone. The stoichiometric coefficients (1 for A, 2 for B) are irrelevant here; only the exponents in the experimentally determined rate law matter.

When we write a rate law like $\text{Rate} = k[A]^m[B]^n$, the exponents $m$ and $n$ describe how sensitive the rate is to changes in each reactant's concentration. The overall order is simply the sum of these individual orders.

Finding the Overall Order

The rate law given is:

Let's identify each component:

1. Order with respect to A: The exponent on $[A]$ is $\frac{1}{2}$. This means the reaction is half-order in A—if you quadruple the concentration of A (keeping B constant), the rate only doubles.

2. Order with respect to B: The exponent on $[B]$ is $1$. This means the reaction is first-order in B—doubling B doubles the rate (keeping A constant).

3. Overall order: Add the individual orders:

   $\text{Overall order} = \frac{1}{2} + 1 = \frac{3}{2} = 1.5$ …

For aA + bB -> products, the rate of reaction is defined so that

Rate = -(1/a) d[A]/dt = -(1/b) d[B]/dt.

Equating the two: (1/a)(-d[A]/dt) = (1/b)(-d[B]/dt), hence

-d[A]/dt = (a/b)(-d[B]/dt) = (-a/b) d[B]/dt. …

Understanding Reaction Rate Stoichiometry

When a chemical reaction proceeds, the concentrations of reactants decrease while products increase. The rate at which each species changes is directly tied to the stoichiometric coefficients in the balanced equation. The key insight is that these rates must be proportional to maintain the stoichiometry throughout the reaction.

For a general reaction $aA + bB \rightarrow cC + dD$, the relationship between the rates of change is:

The negative signs appear for reactants (concentrations decreasing), positive for products (concentrations increasing). Dividing by the stoichiometric coefficient normalizes the rate to a common "reaction rate."

Step-by-Step Analysis

1. Identify the stoichiometry

   For $X + 2Y \rightarrow P$, we have:

   • 1 mole of $X$ consumed
   • 2 moles of $Y$ consumed
   • 1 mole of $P$ formed

2. Write the general rate relationship

   Applying the stoichiometric rate law:

Each term equals the overall reaction rate. Notice $Y$ has coefficient 2, so its rate of change is divided by 2.

3. Extract pairwise relationships

   From the equality above, we can derive several equivalent statements:

   • Between $Y$ and $P$: $-\frac{1}{2}\frac{d[Y]}{dt} = +\frac{d[P]}{dt}$

   Multiplying both sides by $-2$:

4. Check each option …

Understanding Rate Laws and Reaction Order

The rate of a reaction tells us how fast a reactant disappears or a product forms. For an elementary reaction, the rate law can be written directly from the stoichiometry. For $P \rightarrow Q$, the rate of disappearance of $P$ is:

where $k$ is the rate constant and $n$ is the order of the reaction with respect to $P$. The order tells us how sensitive the rate is to changes in concentration.

The key insight here is that if we know how the rate changes when concentration changes, we can deduce the order. Specifically, if concentration is multiplied by some factor, the rate is multiplied by that factor raised to the power of the order.

Finding the Order Step by Step

1. Write the rate law at the initial concentration Let the initial concentration of $P$ be $[P]_0$. The initial rate is:

2. Write the rate law at the doubled concentration When we double the concentration, $[P] = 2[P]_0$. The new rate is:

3. Set up the ratio of rates We're told that the rate increases by a factor of 8, meaning $\text{Rate}_2 = 8 \times \text{Rate}_1$. Taking the ratio:

4. Solve for the order …

Understanding Reaction Rate Stoichiometry

When a chemical reaction proceeds, different species appear or disappear at rates determined by the balanced equation. The stoichiometric coefficients tell us the ratio in which molecules react and form, but the actual rates of change of concentration must account for these ratios.

For the reaction $3A \rightarrow 2B$, three molecules of $A$ disappear for every two molecules of $B$ that form. This means $A$ is consumed $\frac{3}{2}$ times faster than $B$ is produced. The key insight: to define a single "rate of reaction" that's the same no matter which species we measure, we divide each species' rate by its stoichiometric coefficient.

The general relationship is:

The negative sign appears for reactants (concentration decreasing) and positive for products (concentration increasing).

Step-by-Step Solution

1. Write the stoichiometric relationship From $3A \rightarrow 2B$, the stoichiometric coefficients are 3 for $A$ and 2 for $B$. The rate of reaction is defined as:

2. Isolate $\frac{d[B]}{dt}$ We want to express $\frac{d[B]}{dt}$ in terms of $\frac{d[A]}{dt}$. From the equality above:

3. Solve for $\frac{d[B]}{dt}$ Multiply both sides by 2: …