For a chemical reaction A $\rightarrow$ B, the rate of reaction doubles when the concentration of A is increased four times. The order of reaction for A is $\frac{1}{2}$.
If the rate of the chemical reaction A $\rightarrow$ B doubles when the concentration of A is increased four times, it implies that the rate of reaction is directly proportional to the concentration of A to the power of some order.
Let's assume the order of the reaction for A is $n$. According to the given information, when the concentration of A is increased four times (A $\rightarrow$ 4A), the rate of the reaction ($r$) doubles ($r \rightarrow 2r$). Mathematically, this can be expressed as:
$r \propto [A]^n$
$(2r) = k \cdot (4A)^n$
Since $(4A)^n$ can be simplified as $4^n \cdot A^n$, the equation becomes:
$2r = k \cdot 4^n \cdot A^n$
Now, let's consider the original condition where the concentration of A is unchanged (A $\rightarrow$ A). In this case, the rate of reaction is denoted as $r_0$.
$r_0 \propto [A]^n$
Since $[A]^n = A^n$, the equation becomes:
$r_0 = k \cdot A^n$
We can compare the equations for $r$ and $r_0$:
$2r = k \cdot 4^n \cdot A^n$
$r_0 = k \cdot A^n$
To find the order of the reaction for A, we can divide the two equations:
$\frac{2r}{r_0} = \frac{k \cdot 4^n \cdot A^n}{k \cdot A^n}$
$2 = 4^n$
Taking the logarithm (base 2) of both sides:
$\log_2(2) = \log_2(4^n)$
$1 = n \cdot \log_2(4)$
$1 = 2n$
Dividing both sides by 2:
$\frac{1}{2} = n$
Therefore, the order of the reaction for A is $\frac{1}{2}$. |
