Consider a reaction A + R - Product. The rate of this reaction is measured to be k[A][R]. At the st

Q: Why is it called a 'pseudo' first-order reaction if there are two reactants?

It's called 'pseudo' (meaning fake or apparent) because the true reaction is second-order overall (first-order with respect to A and first-order with respect to R). However, because R is in such massive excess, its concentration barely changes. To an observer, the rate appears to depend only on the changing concentration of A, making it look like a first-order reaction.

Q: Why did the concentration of R change to $9.6A_0$? Shouldn't it just be $10A_0$ minus $40\%$?

This is a common trap! The reaction is $40\%$ complete with respect to the limiting reactant, A. This means $0.4A_0$ moles of A have reacted. Because the stoichiometry is 1:1 ($A + R \rightarrow \text{Product}$), exactly $0.4A_0$ moles of R must also react. So, we subtract $0.4A_0$ from the initial $10A_0$ to get $9.6A_0$.

Q: Why do we divide by the True Rate and not the Assumed Rate when calculating the error?

Relative error is a measure of how far off an approximation is from the absolute truth. Therefore, the denominator must always be the exact, true value. Dividing by the assumed value would give you the error relative to the fake scenario, which isn't physically meaningful.

Q: Does the value of the rate constant 'k' affect the final percentage error?

No, it does not! As you can see in the final calculation step, the rate constant 'k' appears in every term in both the numerator and the denominator. It factors out and cancels completely. The error depends purely on the ratio of the initial excess.

Q: What would happen to the error if the reaction was $90\%$ complete instead of $40\%$?

The error would increase! At $90\%$ completion, $0.9A_0$ of R would have reacted, leaving $9.1A_0$. The error would be $\frac{10 - 9.1}{9.1} \times 100 \approx 9.89\%$. As the reaction proceeds further, the assumption that R is constant becomes less and less accurate.

Question

Consider a reaction $A + R \rightarrow Product$. The rate of this reaction is measured to be $k[A][R]$. At the start of the reaction, the concentration of $R$, $[R]_0$, is 10-times the concentration of $A$, $[A]_0$. The reaction can be considered to be a pseudo first order reaction with assumption that $k[R] = k'$ is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40% complete, is ______.
[$k$ and $k'$ represent corresponding rate constants]

Sigma Percentile · Accepted Answer

### The Illusion of Constancy

Welcome to a fascinating exploration of chemical kinetics! In the world of chemistry, we often encounter reactions where one reactant is present in such a massive excess that it seems to defy the laws of consumption. This is the realm of the **pseudo first-order reaction**.

Imagine a reaction where two molecules, $A$ and $R$, must collide to form a product. The rate of this reaction is fundamentally governed by the concentration of both reactants, expressed mathematically as $	ext{Rate} = k[A][R]$.

However, in our specific scenario, the initial concentration of $R$, denoted as $[R]_0$, is a staggering **10 times** the initial concentration of $A$, $[A]_0$. Because $R$ is so abundant, even as the reaction proceeds and consumes $A$, the overall amount of $R$ barely registers a dent. This leads chemists to make a powerful, simplifying assumption: we pretend that the concentration of $R$ remains perfectly constant throughout the reaction. But is this assumption flawless? Let's find out.

### Unmasking the True Rate

To understand the cost of our assumption, we must first calculate the absolute truth. The problem asks us to evaluate the situation when the reaction is **$40\%$ complete**.

If the reaction is $40\%$ complete, it means $40\%$ of our limiting reactant, $A$, has been consumed. The remaining concentration of $A$ is simply:

$$[A] = A_0 - 0.4A_0 = 0.6A_0$$

Now, chemistry is a game of strict stoichiometry. For every molecule of $A$ that reacts, exactly one molecule of $R$ must also react. Therefore, if $0.4A_0$ of $A$ has vanished, exactly $0.4A_0$ of $R$ has also been consumed. The true remaining concentration of $R$ is:

$$[R] = 10A_0 - 0.4A_0 = 9.6A_0$$

Plugging these precise values into our fundamental rate law gives us the **True Rate** ($Rate_1$):

$$Rate_1 = k(0.6A_0)(9.6A_0)$$

### The Pseudo Assumption

Now, let's put on our "pseudo" glasses. The assumption dictates that because $R$ is in such high excess, we can ignore its slight depletion. We assume its concentration remains locked at its initial value:

$$[R]_{	ext{assumed}} = 10A_0$$

Under this assumption, the rate law morphs into a pseudo first-order expression, $	ext{Rate} = k'[A]$, where $k' = k[R]_0$. At $40\%$ completion, the concentration of $A$ is still $0.6A_0$, but we use our assumed, unchanging value for $R$. This gives us the **Assumed Rate** ($Rate_2$):

$$Rate_2 = k(10A_0)(0.6A_0)$$

### Calculating the Price of Approximation

Every approximation carries a cost, and in science, we measure this cost as **Relative Error**. The formula for relative error compares the deviation of our assumed value from the true value, relative to the true value itself:

$$	ext{Relative Error } (\%) = \frac{|	ext{Assumed Rate} - 	ext{True Rate}|}{	ext{True Rate}} 	imes 100$$

Let's substitute our expressions into this formula:

$$	ext{Error} = \frac{k(10A_0)(0.6A_0) - k(9.6A_0)(0.6A_0)}{k(9.6A_0)(0.6A_0)} 	imes 100$$

Here is where the math becomes beautiful. Notice how the rate constant $k$, the initial concentration $A_0$, and the remaining fraction of $A$ ($0.6$) are present in every single term. They factor out and cancel completely! We are left with a pure, elegant ratio of the concentrations of $R$:

$$	ext{Error} = \frac{10 - 9.6}{9.6} 	imes 100$$

$$	ext{Error} = \frac{0.4}{9.6} 	imes 100 = 4.166... \%$$

Rounding to two decimal places, we arrive at our final answer: **$4.17\%$**.

### The Grand Takeaway

An error of roughly $4.2\%$ is generally acceptable in many experimental setups, validating the utility of the pseudo first-order assumption. However, this problem teaches us a deeper lesson.

Imagine if the initial concentration of $R$ was $100$ times that of $A$. The remaining $R$ would be $99.6A_0$, and the error would plummet to just $0.4\%$. The core takeaway is profound: **The validity of a pseudo-order approximation scales directly with the magnitude of the excess.** The larger the excess, the closer the illusion of constancy approaches reality.

Enter Numerical Value:

Visualized Solution

$A + R \to Product$

$True Rate Law$

$True Rate Calculation$

$Pseudo First-Order Assumption$

$Assumed Rate Calculation$

$Relative Error Formula$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Order and Molecularity

The Illusion of Constancy

Unmasking the True Rate

The Pseudo Assumption

Calculating the Price of Approximation

The Grand Takeaway

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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