For the non-stoichiometric reaction, 2A + B - C + D, the following kinetic data were obtained in th

Q: Why do we assume the rate law is $\text{Rate} = k[A]^x[B]^y$?

This is the standard empirical form for a rate law. The exponents $x$ and $y$ represent the order of the reaction with respect to each reactant, which must be determined experimentally. They are not necessarily related to the stoichiometric coefficients.

Q: Can we just use the stoichiometric coefficients as the powers?

No! For a non-elementary reaction, the rate law cannot be predicted from the balanced chemical equation. The powers must be found using experimental data. Only for strictly elementary (single-step) reactions do the coefficients match the orders.

Q: What does it mean when the order is zero?

A zero-order with respect to a reactant means that changing its concentration has absolutely no effect on the rate of the reaction, as long as some of it is present. It usually implies that the reactant is involved in a fast step after the slow, rate-determining step.

Q: Why did we divide the rate equations?

Dividing the rate equations of two different experiments allows us to cancel out the rate constant $k$ and the concentration of the reactant that was kept constant. This leaves us with a simple equation containing only one unknown exponent, making it easy to solve.

Q: What if no two experiments have a constant concentration for one reactant?

In that case, you would first determine the order for one reactant using a pair of experiments where the other is constant. Then, use that known order in another pair of experiments to solve for the remaining unknown order algebraically.

Question

For the non-stoichiometric reaction, $2A + B ightarrow C + D$, the following kinetic data were obtained in three separate experiments, all at $298	ext{ K}$.

$\begin{array}{|c|c|c|}\hline 	ext{Initial concentration (A)} & 	ext{Initial concentration (B)} & 	ext{Initial rate of formation of C (mol L}^{-1}	ext{s}^{-1}	ext{)} \ \hline 0.1 	ext{ M} & 0.1 	ext{ M} & 1.2 	imes 10^{-3} \ \hline 0.1 	ext{ M} & 0.2 	ext{ M} & 1.2 	imes 10^{-3} \ \hline 0.2 	ext{ M} & 0.1 	ext{ M} & 2.4 	imes 10^{-3} \ \hline \end{array}$

The rate law for the formation of C is

Options:

(A) $\frac{dc}{dt} = k[A][B]$

(B) $\frac{dc}{dt} = k[A]^2[B]$

(C) $\frac{dc}{dt} = k[A][B]^2$

(D) $\frac{dc}{dt} = k[A]$

Sigma Percentile · Accepted Answer

### Unraveling the Rate Law using Initial Rates

When dealing with chemical kinetics, one of the most fundamental tasks is determining the rate law of a reaction. The rate law tells us exactly how the speed of the reaction depends on the concentration of each reactant.

For the given non-stoichiometric reaction, $2A + B ightarrow C + D$, we cannot simply look at the balanced equation and assume the stoichiometric coefficients are the powers in the rate law. Instead, we must rely on experimental data. We start by writing a general empirical rate law:

$$ 	ext{Rate} = k[A]^x[B]^y $$

Here, $k$ is the rate constant, and $x$ and $y$ are the unknown orders of the reaction with respect to reactants $A$ and $B$, respectively.

### Isolating the Effect of B

To find these unknown powers, we use the **Method of Initial Rates**. The core logic is simple: if we want to see how reactant $B$ affects the rate, we must find two experiments where the concentration of $A$ is kept perfectly constant while the concentration of $B$ changes.

Looking at the data table, Experiments 1 and 2 fit this criteria perfectly. In both experiments, $[A]$ is held constant at $0.1	ext{ M}$. However, $[B]$ is doubled from $0.1	ext{ M}$ to $0.2	ext{ M}$.

Let's observe what happens to the rate. The rate in Experiment 1 is $1.2 	imes 10^{-3}	ext{ M/s}$, and in Experiment 2, it is also $1.2 	imes 10^{-3}	ext{ M/s}$. Even though we doubled the amount of $B$, the rate didn't budge! Mathematically, we can express this by dividing the two rate equations:

$$ \frac{	ext{Rate}_2}{	ext{Rate}_1} = \frac{k(0.1)^x(0.2)^y}{k(0.1)^x(0.1)^y} $$

$$ 1 = 2^y $$

The only way $2^y$ can equal $1$ is if $y = 0$. This means the reaction is **zero-order** with respect to $B$.

### Isolating the Effect of A

Now, we apply the exact same logic to find $x$. We need two experiments where $[B]$ is constant, but $[A]$ changes. Experiments 1 and 3 are our perfect match. Here, $[B]$ is held at $0.1	ext{ M}$, while $[A]$ is doubled from $0.1	ext{ M}$ to $0.2	ext{ M}$.

What happens to the rate this time? It jumps from $1.2 	imes 10^{-3}	ext{ M/s}$ to $2.4 	imes 10^{-3}	ext{ M/s}$. The rate has exactly doubled! Let's set up the math:

$$ \frac{	ext{Rate}_3}{	ext{Rate}_1} = \frac{k(0.2)^x(0.1)^y}{k(0.1)^x(0.1)^y} $$

$$ 2 = 2^x $$

For this to be true, $x$ must equal $1$. The reaction is **first-order** with respect to $A$.

### The Final Rate Expression

We have successfully decoded the experimental data. By substituting $x = 1$ and $y = 0$ back into our general rate law, we get:

$$ 	ext{Rate} = k[A]^1[B]^0 $$

Since anything raised to the power of zero is one, the term for $[B]$ drops out entirely. Our final, elegant rate law is simply:

$$ \frac{dc}{dt} = k[A] $$

This tells us that the speed of this reaction is entirely dictated by the concentration of $A$, and $B$ is merely a silent participant in the rate-determining step.

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Visualized Solution

$The General Rate Law$

$Method of Initial Rates$

$Isolating B$

$Order with respect to B$

$Isolating A$

$Order with respect to A$

$Final Rate Law$

The Sigma Insight: Rate of Chemical Reaction

Unraveling the Rate Law using Initial Rates

Isolating the Effect of B

Isolating the Effect of A

The Final Rate Expression

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The General Rate Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Method of Initial Rates

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Order with respect to B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Order with respect to A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Rate Law