Consider the kinetic data given in the following table for the reaction A + B + C - Product.
Experi

Q: Why did we completely ignore the concentration of B in the final calculation?

Because we found that the reaction is zero-order with respect to B ($n_2 = 0$). Any number raised to the power of zero is 1. This means the rate is completely independent of B's concentration, so changing it has no effect on the reaction speed.

Q: Could we have used Experiment 2 or 3 to find the rate constant K instead of Experiment 1?

Absolutely! The rate constant $K$ is a constant for a given reaction at a specific temperature. Using the data from any of the four experiments will give you the exact same value of $3 \times 10^{-3}$.

Q: What if the rate didn't increase by an exact integer multiple, like doubling or tripling?

You would use logarithms. For example, if $\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{A}]_2}{[\text{A}]_1}\right)^n$, you can take the log of both sides to solve for $n$: $n = \frac{\log(\text{Rate}_2 / \text{Rate}_1)}{\log([\text{A}]_2 / [\text{A}]_1)}$.

Q: Why are the units of K not mentioned in the final calculation?

While $K$ does have specific units (in this case, $\text{mol}^{-1}\text{dm}^3\text{s}^{-1}$ for a second-order overall reaction), the question only asks for the numerical value of $Y$. As long as you are consistent with your concentration and time units throughout the calculation, the final numerical value will be correct.

Q: Is it possible for a reaction order to be negative or a fraction?

Yes! A negative order means the reactant acts as an inhibitor, slowing down the reaction as its concentration increases. Fractional orders often occur in complex, multi-step reaction mechanisms. However, in simple initial rate problems like this, they are usually integers.

Question

Consider the kinetic data given in the following table for the reaction $	ext{A} + 	ext{B} + 	ext{C} ightarrow 	ext{Product}$.

$\begin{array}{|c|c|c|c|c|}
\hline
	ext{Experiment No.} & 	ext{[A]} (	ext{mol dm}^{-3}) & 	ext{[B]} (	ext{mol dm}^{-3}) & 	ext{[C]} (	ext{mol dm}^{-3}) & 	ext{Rate of reaction} (	ext{mol dm}^{-3}	ext{s}^{-1}) \
\hline
1 & 0.2 & 0.1 & 0.1 & 6.0 	imes 10^{-5} \
\hline
2 & 0.2 & 0.2 & 0.1 & 6.0 	imes 10^{-5} \
\hline
3 & 0.2 & 0.1 & 0.2 & 1.2 	imes 10^{-4} \
\hline
4 & 0.3 & 0.1 & 0.1 & 9.0 	imes 10^{-5} \
\hline
\end{array}$

The rate of the reaction for $	ext{[A]} = 0.15 	ext{ mol dm}^{-3}$, $	ext{[B]} = 0.25 	ext{ mol dm}^{-3}$ and $	ext{[C]} = 0.15 	ext{ mol dm}^{-3}$ is found to be $Y 	imes 10^{-5} 	ext{ mol dm}^{-3}	ext{s}^{-1}$. The value of $Y$ is ________.

Sigma Percentile · Accepted Answer

Imagine you are a detective trying to figure out which suspects are actually involved in a crime. In chemical kinetics, the 'crime' is the reaction happening, and the 'suspects' are the reactants. Some reactants drive the reaction forward aggressively, while others might just be standing around doing nothing. The **Method of Initial Rates** is our interrogation technique.

### Analyzing the Setup

We are given a reaction with three reactants: A, B, and C. The general rate law is our starting hypothesis:

$$r = K[	ext{A}]^{n_1}[	ext{B}]^{n_2}[	ext{C}]^{n_3}$$

Our mission is to find the orders $n_1$, $n_2$, and $n_3$. To do this, we look at the experimental data and find pairs of experiments where only *one* reactant's concentration changes. This isolates its effect on the rate.

### Interrogating Reactant B

Let's compare Experiment 1 and Experiment 2. Notice how the concentrations of A and C are kept perfectly constant at $0.2$ and $0.1$ respectively. However, the concentration of B is doubled from $0.1$ to $0.2$.

What happens to the rate? Absolutely nothing! It stays at $6.0 	imes 10^{-5}$. This means reactant B is a silent spectator. It has zero effect on the speed of the reaction. Therefore, the order with respect to B is zero ($n_2 = 0$).

### Interrogating Reactant C

Next, we compare Experiment 1 and Experiment 3. This time, A and B are constant, but C is doubled from $0.1$ to $0.2$.

The rate jumps from $6.0 	imes 10^{-5}$ to $1.2 	imes 10^{-4}$. That is exactly double! Because the rate scales linearly with the concentration of C, the reaction is first order with respect to C ($n_3 = 1$).

### Interrogating Reactant A

Finally, let's look at Experiment 1 and Experiment 4. The concentration of A increases by a factor of 1.5 (from $0.2$ to $0.3$).

The rate also increases by a factor of 1.5 (from $6.0 	imes 10^{-5}$ to $9.0 	imes 10^{-5}$). Once again, we see a direct, linear relationship. The order with respect to A is also one ($n_1 = 1$).

### The Master Equation and Final Calculation

Now we can write the true rate law:

$$r = K[	ext{A}][	ext{C}]$$

Before we can find the final answer, we need the rate constant, $K$. We can use the data from any experiment to find it. Let's use Experiment 1:

$$6.0 	imes 10^{-5} = K(0.2)(0.1)$$

$$K = \frac{6.0 	imes 10^{-5}}{0.02} = 3 	imes 10^{-3}$$

Now for the grand finale. The question asks for the rate when $[	ext{A}] = 0.15$, $[	ext{B}] = 0.25$, and $[	ext{C}] = 0.15$. Remember, B is a spectator, so we ignore its concentration entirely!

$$r = (3 	imes 10^{-3})(0.15)(0.15)$$

$$r = 6.75 	imes 10^{-5}$$

Comparing this to the given format $Y 	imes 10^{-5}$, we find that **$Y = 6.75$**. A beautiful, logical deduction from start to finish!

Enter Numerical Value:

Visualized Solution

$r = K [A]^{n_{1}} [B]^{n_{2}} [C]^{n_{3}}$

$Order of B$

$Order of C$

$Order of A$

$Rate Law & K$

$Calculate K$

$Final Substitution$

$Final Answer$

The Sigma Insight: Order and Molecularity

Analyzing the Setup

Interrogating Reactant B

Interrogating Reactant C

Interrogating Reactant A

The Master Equation and Final Calculation

Similar Questions

Consider following two reactions, $A ⟶ Product; - \frac{d [ A ]}{d t} = k_{1} [A]^{0}$ $B ⟶ Product; - \frac{d [ B ]}{d t} = k_{2} [B]$ $k_{1}$ and $k_{2}$ are expressed in terms of molarity ( $mol L^{- 1}$ ) and time ( $s^{- 1}$ ) as

Consider the following reactions $A \to P 1; B \to P 2; C \to P 3; D \to P 4$ , The order of the above reactions are $a, b, c$ and $d$ , respectively. The following graph is obtained when $lo g [rate]$ vs $lo g [conc.]$ are plotted :

For a reaction $A + 2 B ⟶ C$ , rate is given by $+ \frac{d [ C ]}{d t} = k [A] [B]$ , hence the order of the reaction is

The given plots represent the variation of the concentration of a reaction $R$ with time for two different reactions (i) and (ii). The respective orders of the reactions are

The rate equation for the reaction $2 A + B \to C$ is found to be rate $= k [A] [B]$ . The correct statement in relation to this reaction is that the

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=K[A]n1​[B]n2​[C]n3​

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

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

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

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

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

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

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