The following data was obtained for chemical reaction given below at 975 K.
$$2NO_{(g)} + 2H_{2(g)} \rightarrow N_{2(g)} + 2H_2O_{(g)}$$
[NO] = 8 $$\times 10^{-5}$$, [H$$_2$$] = 8 $$\times 10^{-5}$$, Rate = 7 $$\times 10^{-9}$$
[NO] = 24 $$\times 10^{-5}$$, [H$$_2$$] = 8 $$\times 10^{-5}$$, Rate = 2.1 $$\times 10^{-8}$$
[NO] = 24 $$\times 10^{-5}$$, [H$$_2$$] = 32 $$\times 10^{-5}$$, Rate = 8.4 $$\times 10^{-8}$$
The order of the reaction with respect to NO is _________ [Integer answer]

For any chemical reaction, we write a differential rate law. For the present reaction $$2\,NO_{(g)} + 2\,H_{2\,(g)} \;\longrightarrow\; N_{2\,(g)} + 2\,H_{2}O_{(g)},$$ let the rate be expressed as

$$\text{Rate}=k\,[NO]^{\,x}\,[H_2]^{\,y}.$$

Here $$k$$ is the rate constant, $$x$$ is the order with respect to $$NO$$, and $$y$$ is the order with respect to $$H_2$$. We have to determine $$x$$ only, so we shall select two experimental runs in which the concentration of $$H_2$$ is kept the same while the concentration of $$NO$$ is changed. That will make all the $$[H_2]$$ factors cancel out in the ratio.

The first and second experiments satisfy this requirement because in both cases $$[H_2]=8\times10^{-5}\;{\rm M}$$, whereas $$[NO]$$ changes.

For experiment 1 $$[NO]_1 = 8\times10^{-5}\;{\rm M}, \qquad \text{Rate}_1 = 7\times10^{-9}\;{\rm M\,s^{-1}}.$$

For experiment 2 $$[NO]_2 = 24\times10^{-5}\;{\rm M}, \qquad \text{Rate}_2 = 2.1\times10^{-8}\;{\rm M\,s^{-1}}.$$

Now we take the ratio of the two rate expressions. First we write them explicitly:

$$ \begin{aligned} \text{Rate}_1 &= k\,[NO]_1^{\,x}\,[H_2]_1^{\,y},\\[4pt] \text{Rate}_2 &= k\,[NO]_2^{\,x}\,[H_2]_2^{\,y}. \end{aligned} $$

Dividing the second equation by the first gives

$$ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k\,[NO]_2^{\,x}\,[H_2]_2^{\,y}}{k\,[NO]_1^{\,x}\,[H_2]_1^{\,y}} = \left(\frac{[NO]_2}{[NO]_1}\right)^{x} \left(\frac{[H_2]_2}{[H_2]_1}\right)^{y}. $$

Because $$[H_2]_1=[H_2]_2,$$ their ratio equals $$1$$, and any power of $$1$$ is still $$1$$, so the $$H_2$$ factor disappears:

$$ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[NO]_2}{[NO]_1}\right)^{x}. $$

Substituting the numerical values, we have

$$ \frac{2.1\times10^{-8}}{7\times10^{-9}} = \left(\frac{24\times10^{-5}}{8\times10^{-5}}\right)^{x}. $$

Simplifying both fractions step by step:

Left-hand side (LHS): $$ \frac{2.1\times10^{-8}}{7\times10^{-9}} = \frac{2.1}{7}\times\frac{10^{-8}}{10^{-9}} = 0.3\times10^{1} = 3. $$

Right-hand side (RHS) inside the parentheses:

$$ \frac{24\times10^{-5}}{8\times10^{-5}} = \frac{24}{8}\times\frac{10^{-5}}{10^{-5}} = 3\times1 = 3. $$

Therefore we get

$$3 = 3^{\,x}.$$

We recognize that $$3^{\,1}=3$$ and $$3^{\,2}=9$$, etc. Hence the only way for $$3^{\,x}$$ to equal $$3$$ is for $$x=1$$.

We can confirm this value quickly with the third experiment. In that case both $$[NO]$$ and $$[H_2]$$ change, but if we already fix $$x=1$$ we can verify internally that a consistent integer value of $$y$$ arises. However, that check is not necessary for the asked quantity; our algebra has already given a unique integer value for the order with respect to $$NO$$.

So, the answer is $$1$$.