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Step-by-Step Solution
Step 1: Write the Arrhenius Relation for Reaction A
For reaction A, the rate doubles when temperature is increased from 300 K to 310 K. We can express this using the Arrhenius equation in logarithmic form:
$ \log\left(\frac{k_2}{k_1}\right) = \frac{E_{a_1}}{2.303 R} \left[\frac{1}{T_1} - \frac{1}{T_2}\right] $
Here,
$k_1$ is the rate constant at $T_1 = 300\,\text{K}$
$k_2$ is the rate constant at $T_2 = 310\,\text{K}$
$E_{a_1}$ is the activation energy for reaction A
$R$ is the gas constant
Given $k_2 = 2 k_1$, substitute this into the equation:
$ \log(2) = \frac{E_{a_1}}{2.303 R} \left[\frac{1}{300} - \frac{1}{310}\right] \;...\;(1)
$
Step 2: Write the Arrhenius Relation for Reaction B
For reaction B, the rate also doubles, but the activation energy is twice that of reaction A. Let $E_{a_2} = 2 E_{a_1}$. The initial temperature remains $T_1 = 300\,\text{K}$, and the new temperature is $T_2 = ?$. Again the rate doubles, so we can write:
$ \log(2) = \frac{E_{a_2}}{2.303 R} \left[\frac{1}{300} - \frac{1}{T_2}\right]
$
Substitute $E_{a_2} = 2 E_{a_1}$:
$ \log(2) = \frac{2 E_{a_1}}{2.303 R} \left[\frac{1}{300} - \frac{1}{T_2}\right] \;...\;(2)
$
Step 3: Relate Equations (1) and (2) to Find $T_2$ for Reaction B
From equation (1), we have:
$ \log(2) = \frac{E_{a_1}}{2.303 R} \left[\frac{1}{300} - \frac{1}{310}\right]
$
From equation (2), we have:
$ \log(2) = \frac{2 E_{a_1}}{2.303 R} \left[\frac{1}{300} - \frac{1}{T_2}\right]
$
Set them equal:
$ \frac{2 E_{a_1}}{2.303 R} \left[\frac{1}{300} - \frac{1}{T_2}\right]
=
\frac{E_{a_1}}{2.303 R} \left[\frac{1}{300} - \frac{1}{310}\right]
$
Cancel out the common factor $ \frac{E_{a_1}}{2.303 R} $:
$ 2 \left[\frac{1}{300} - \frac{1}{T_2}\right]
=
\left[\frac{1}{300} - \frac{1}{310}\right]
$
Step 4: Solve for $T_2$
Rearranging the above equation:
$ \frac{1}{300} - \frac{1}{T_2}
= \frac{1}{2} \left[\frac{1}{300} - \frac{1}{310}\right]
$
Compute the bracket on the right side:
$ \frac{1}{300} - \frac{1}{310}
= \frac{310 - 300}{300 \times 310}
= \frac{10}{300 \times 310}
= \frac{10}{93000}
= \frac{1}{9300}
$
So,
$ \frac{1}{300} - \frac{1}{T_2}
= \frac{1}{2} \times \frac{1}{9300}
= \frac{1}{18600}
$
Now,
$ \frac{1}{T_2}
= \frac{1}{300} - \frac{1}{18600}
$
Find a common denominator or directly simplify. However, an equivalent step given in the provided solution uses a more direct approach:
$ T_2 = 304.92\,\text{K} \,(\text{approximately})
$
Step 5: Calculate the Increase in Temperature
Since the original temperature was 300 K:
$ \Delta T = T_2 - 300 = 304.92 - 300 = 4.92\,\text{K}
$
Therefore, the temperature must be increased by about 4.92 K to double the rate for reaction B.
Answer: 4.92 K
