The following data was obtained for chemical reaction given below at 975 K.
2NO(g) + 2H2(g) - N2(g)

Q: Why do we divide the equations instead of subtracting them?

Great question! The rate law involves multiplication ($K \times [\text{NO}]^x \times [\text{H}_2]^y$). To eliminate terms that are multiplied, we must use division. Subtracting them wouldn't cancel out the unknown rate constant $K$ or the $[\text{H}_2]^y$ term.

Q: Can I use experiments A and C to find the order of NO?

You could, but it would be much harder! In experiments A and C, both the NO and $\text{H}_2$ concentrations change. If you divide their equations, you'll be left with an equation containing both $x$ and $y$ as unknowns, which you can't solve without finding $y$ first.

Q: Why did we change $2.1 \times 10^{-8}$ to $21 \times 10^{-9}$?

This is a handy math trick! By making the powers of 10 identical in the numerator and denominator (both $10^{-9}$), we can easily cancel them out. It turns a messy decimal division into a simple fraction: $7/21$, which is exactly $1/3$.

Q: What if the concentrations were never constant in any two experiments?

That's a tricky scenario! You would have to divide two equations to get a relationship between $x$ and $y$, and then divide another pair to get a second relationship. You'd then solve them as a system of simultaneous equations.

Q: Does the temperature of 975 K matter for this specific calculation?

Not directly for finding the order! The temperature is given to assure you that the rate constant $K$ remains the same across all three experiments. If the temperature changed between experiments, $K$ would change, and we wouldn't be able to cancel it out!

Question

The following data was obtained for chemical reaction given below at 975 K.
$2	ext{NO}(g) + 2	ext{H}_2(g) ightarrow 	ext{N}_2(g) + 2	ext{H}_2	ext{O}(g)$
$\begin{array}{lccc} \hline & 	ext{[NO]} & 	ext{[H}_2	ext{]} & 	ext{[Rate]} \ & 	ext{mol L}^{-1} & 	ext{mol L}^{-1} & 	ext{mol L}^{-1} \ \hline 	ext{A.} & 8 	imes 10^{-5} & 8 	imes 10^{-5} & 7 	imes 10^{-9} \ 	ext{B.} & 24 	imes 10^{-5} & 8 	imes 10^{-5} & 2.1 	imes 10^{-8} \ 	ext{C.} & 24 	imes 10^{-5} & 32 	imes 10^{-5} & 8.4 	imes 10^{-8} \ \hline \end{array}$
The order of the reaction with respect to NO is \dots\dots . [Integer answer]

Sigma Percentile · Accepted Answer

### The Master Equation

Imagine you are a detective trying to solve a mystery. In chemical kinetics, the mystery is often figuring out exactly how the concentration of each reactant affects the speed of the reaction.

To start our investigation, we write down the general rate law. This is our master equation.

$$ 	ext{Rate} = K[	ext{NO}]^x[	ext{H}_2]^y $$

Here, $K$ is the rate constant, and $x$ and $y$ are the orders of the reaction with respect to nitric oxide and hydrogen gas, respectively. Our primary mission is to find the value of $x$.

### The Art of Elimination

To find the order with respect to NO, which is $x$, we need a clever trick. If we look at the rate law, we have two unknowns in the powers: $x$ and $y$.

We need to isolate $x$. We can do this by finding two experiments where the concentration of hydrogen gas remains exactly the same.

By keeping $[	ext{H}_2]$ constant, its effect on the rate becomes constant, allowing us to cancel it out completely when we compare the two experiments.

### The Mathematical Magic

Let's look closely at the data table provided. Notice experiments A and B?

In both of these trials, the concentration of $	ext{H}_2$ is exactly $8 	imes 10^{-5} 	ext{ mol L}^{-1}$. This is perfect for our elimination strategy!

Now, let's plug the values from row A and row B into our rate law.

For experiment A:
$$ 7 	imes 10^{-9} = K(8 	imes 10^{-5})^x(8 	imes 10^{-5})^y $$

For experiment B:
$$ 2.1 	imes 10^{-8} = K(24 	imes 10^{-5})^x(8 	imes 10^{-5})^y $$

Here comes the magic step. Let's divide the equation for A by the equation for B.

$$ \frac{7 	imes 10^{-9}}{2.1 	imes 10^{-8}} = \frac{K(8 	imes 10^{-5})^x(8 	imes 10^{-5})^y}{K(24 	imes 10^{-5})^x(8 	imes 10^{-5})^y} $$

Watch what happens. The rate constant $K$ cancels out. Beautifully, the entire $	ext{H}_2$ term cancels out as well!

### The Final Verdict

Let's simplify the numbers to make the calculation easier.

We can rewrite $2.1 	imes 10^{-8}$ as $21 	imes 10^{-9}$. Now, the powers of ten cancel out nicely on the left side.

$$ \frac{7 	imes 10^{-9}}{21 	imes 10^{-9}} = \left(\frac{8 	imes 10^{-5}}{24 	imes 10^{-5}}ight)^x $$

This simplifies down to a very clean equation:

$$ \frac{1}{3} = \left(\frac{1}{3}ight)^x $$

Comparing the powers on both sides, it is crystal clear that $x$ must be equal to $1$.

**The order of the reaction with respect to NO is 1.**

Enter Numerical Value:

Visualized Solution

General Rate Law

Strategy to Find $x$

Selecting Experiments

Writing Rate Equations

Dividing Equations

Simplifying the Ratio

Final Calculation

The Way Forward

The Sigma Insight: Order and Molecularity

The Master Equation

The Art of Elimination

The Mathematical Magic

The Final Verdict

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