228209
For a reaction of type $A+B \rightarrow$ products, it is observed that doubling concentration of $A$ causes the reaction rate to be four times as great, but doubling amount of $B$ does not affect the rate. The unit of rate constant is
For a reaction - $\mathrm{A}+\mathrm{B} \rightarrow \text { product }$ Let the initial rate be $\mathrm{R}$ And order with respect to $\mathrm{A}$ be $\mathrm{x}$ and $\mathrm{B}$ be $\mathrm{y}$. Thus, rate law can be written as, Rate, $\mathrm{R}=[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ After doubling the concentration of $\mathrm{A}$, rate becomes $4 \mathrm{R}$, $4 \mathrm{R}=[2 \mathrm{~A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ After doubling the concentration of $\mathrm{B}$, rate remains $\mathrm{R}$, $\mathrm{R}=[\mathrm{A}]^{\mathrm{x}}[2 \mathrm{~B}]^{\mathrm{y}}$ From equation (i) and (ii), we get $\frac{\mathrm{R}}{4 \mathrm{R}}=\left(\frac{1}{2}\right)^{\mathrm{x}} \Rightarrow\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{\mathrm{x}}$ So, $\quad \mathrm{x}=2$ From equation (i) and (iii), we get $\frac{\mathrm{R}}{\mathrm{R}}=\left[\frac{1}{2}\right]^{\mathrm{y}} \Rightarrow\left(\frac{1}{1}\right)^0=\left(\frac{1}{2}\right)^{\mathrm{y}}$ So, $\quad \mathrm{Y}=0$ Hence, the rate law is, rate $\mathrm{R}=[\mathrm{A}]^2[\mathrm{~B}]^0$ This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are $\mathrm{mol}^{-1} \mathrm{Ls}^{-1}$.
228209
For a reaction of type $A+B \rightarrow$ products, it is observed that doubling concentration of $A$ causes the reaction rate to be four times as great, but doubling amount of $B$ does not affect the rate. The unit of rate constant is
For a reaction - $\mathrm{A}+\mathrm{B} \rightarrow \text { product }$ Let the initial rate be $\mathrm{R}$ And order with respect to $\mathrm{A}$ be $\mathrm{x}$ and $\mathrm{B}$ be $\mathrm{y}$. Thus, rate law can be written as, Rate, $\mathrm{R}=[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ After doubling the concentration of $\mathrm{A}$, rate becomes $4 \mathrm{R}$, $4 \mathrm{R}=[2 \mathrm{~A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ After doubling the concentration of $\mathrm{B}$, rate remains $\mathrm{R}$, $\mathrm{R}=[\mathrm{A}]^{\mathrm{x}}[2 \mathrm{~B}]^{\mathrm{y}}$ From equation (i) and (ii), we get $\frac{\mathrm{R}}{4 \mathrm{R}}=\left(\frac{1}{2}\right)^{\mathrm{x}} \Rightarrow\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{\mathrm{x}}$ So, $\quad \mathrm{x}=2$ From equation (i) and (iii), we get $\frac{\mathrm{R}}{\mathrm{R}}=\left[\frac{1}{2}\right]^{\mathrm{y}} \Rightarrow\left(\frac{1}{1}\right)^0=\left(\frac{1}{2}\right)^{\mathrm{y}}$ So, $\quad \mathrm{Y}=0$ Hence, the rate law is, rate $\mathrm{R}=[\mathrm{A}]^2[\mathrm{~B}]^0$ This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are $\mathrm{mol}^{-1} \mathrm{Ls}^{-1}$.
228209
For a reaction of type $A+B \rightarrow$ products, it is observed that doubling concentration of $A$ causes the reaction rate to be four times as great, but doubling amount of $B$ does not affect the rate. The unit of rate constant is
For a reaction - $\mathrm{A}+\mathrm{B} \rightarrow \text { product }$ Let the initial rate be $\mathrm{R}$ And order with respect to $\mathrm{A}$ be $\mathrm{x}$ and $\mathrm{B}$ be $\mathrm{y}$. Thus, rate law can be written as, Rate, $\mathrm{R}=[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ After doubling the concentration of $\mathrm{A}$, rate becomes $4 \mathrm{R}$, $4 \mathrm{R}=[2 \mathrm{~A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ After doubling the concentration of $\mathrm{B}$, rate remains $\mathrm{R}$, $\mathrm{R}=[\mathrm{A}]^{\mathrm{x}}[2 \mathrm{~B}]^{\mathrm{y}}$ From equation (i) and (ii), we get $\frac{\mathrm{R}}{4 \mathrm{R}}=\left(\frac{1}{2}\right)^{\mathrm{x}} \Rightarrow\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{\mathrm{x}}$ So, $\quad \mathrm{x}=2$ From equation (i) and (iii), we get $\frac{\mathrm{R}}{\mathrm{R}}=\left[\frac{1}{2}\right]^{\mathrm{y}} \Rightarrow\left(\frac{1}{1}\right)^0=\left(\frac{1}{2}\right)^{\mathrm{y}}$ So, $\quad \mathrm{Y}=0$ Hence, the rate law is, rate $\mathrm{R}=[\mathrm{A}]^2[\mathrm{~B}]^0$ This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are $\mathrm{mol}^{-1} \mathrm{Ls}^{-1}$.
228209
For a reaction of type $A+B \rightarrow$ products, it is observed that doubling concentration of $A$ causes the reaction rate to be four times as great, but doubling amount of $B$ does not affect the rate. The unit of rate constant is
For a reaction - $\mathrm{A}+\mathrm{B} \rightarrow \text { product }$ Let the initial rate be $\mathrm{R}$ And order with respect to $\mathrm{A}$ be $\mathrm{x}$ and $\mathrm{B}$ be $\mathrm{y}$. Thus, rate law can be written as, Rate, $\mathrm{R}=[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ After doubling the concentration of $\mathrm{A}$, rate becomes $4 \mathrm{R}$, $4 \mathrm{R}=[2 \mathrm{~A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$ After doubling the concentration of $\mathrm{B}$, rate remains $\mathrm{R}$, $\mathrm{R}=[\mathrm{A}]^{\mathrm{x}}[2 \mathrm{~B}]^{\mathrm{y}}$ From equation (i) and (ii), we get $\frac{\mathrm{R}}{4 \mathrm{R}}=\left(\frac{1}{2}\right)^{\mathrm{x}} \Rightarrow\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{\mathrm{x}}$ So, $\quad \mathrm{x}=2$ From equation (i) and (iii), we get $\frac{\mathrm{R}}{\mathrm{R}}=\left[\frac{1}{2}\right]^{\mathrm{y}} \Rightarrow\left(\frac{1}{1}\right)^0=\left(\frac{1}{2}\right)^{\mathrm{y}}$ So, $\quad \mathrm{Y}=0$ Hence, the rate law is, rate $\mathrm{R}=[\mathrm{A}]^2[\mathrm{~B}]^0$ This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are $\mathrm{mol}^{-1} \mathrm{Ls}^{-1}$.