Answer:
![\begin{gathered} a)\text{ Rate = k }*\text{ \lbrack NO\rbrack}^2\text{ }*\text{ \lbrack Cl}_2] \\ b)\text{ Order = 3} \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/e9r8c3fnlw3me4m7b3pgczuzc2qjyz6kdf.png)
Step-by-step explanation:
Here, we want to deduce the rate law for the reaction given
According to the data provided:
![Rate\text{ = k\lbrack NO\rbrack}^a[Cl_2]\placeholder{⬚}^b](https://img.qammunity.org/2023/formulas/chemistry/college/8mazpi06c4f00twtnkgr69haga9mz7ml00.png)
where the values in the square parentheses represent the concentrations and k represents the rate constant
Let us work with equations 1 and 3:
![\begin{gathered} 2.27\text{ }*10^(-5)\text{ = k }*\text{ 0.0125}^a\text{ }*\text{ 0.0255}^b \\ 9.08\text{ }*\text{ 10}^(-5)\text{ = k }*\text{ 0.0250}^a\text{ }*\text{ 0.0255}^b \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/f1rok81bi0kykkuww0zz8zho2evoyq23oi.png)
Divide equation 3 by 1:
![\begin{gathered} 4\text{ = 2}^a \\ a\text{ = 2} \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/q5foid59vb9yszlvp4c604lb3hxnvmgds7.png)
To get b, we can use equations 1 and 2:
![\begin{gathered} 2.27\text{ }*\text{ 10}^(-5)\text{ = k }*\text{ 0.0125}^a\text{ }*\text{ 0.0255}^b \\ 4.55\text{ }*\text{ 10}^(-5)\text{ = k }*\text{ 0.0125}^a\text{ }*\text{ 0.0510}^b \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/3ivgqgxlng7zyklyy5y46eb1xx09n7vn2z.png)
Divide equation 2 by 1, we have it that:
![\begin{gathered} 2\text{ = 2}^b \\ b\text{ = 1} \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/uztcxcabsuwwvbelchytpl23vwrzwvyntl.png)
The rate law for the reaction is thus:
![Rate\text{ = k }*\text{ \lbrack NO\rbrack}^2\text{ }*\text{ \lbrack Cl}_2]](https://img.qammunity.org/2023/formulas/chemistry/college/w0c8dol7x1mgclu70spi7afeqimixgo5kr.png)
b) The overall order is the sum of the powers
That would be 1 + 2 = 3