The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every `10^(@)C` rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every `10^(@)C` change in temperature.
`"Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C)`
Arrhenius gave an equation which describes aret constant `k` as a function of temperature
`k = Ae^(-E_(a)//RT)`
where `k` is the rate constant, `A` is the frequency factor or pre-exponential factor, `E_(a)` is the activation energy, `T` is the temperature in kelvin, `R` is the universal gas constant.
Equation when expressed in logarithmic form becomes
`log k = log A - (E_(a))/(2.303 RT)`
For the given reactions, following data is given
`{:(PrarrQ,,,,k_(1) =10^(15)exp((-2000)/(T))),(CrarrD,,,,k_(2) = 10^(14)exp((-1000)/(T))):}`
Temperature at which `k_(1) = k_(2)` is

Options

A. `434.22 K`
B. `1000 K`
C. `2000 K`
D. `868.44 K`

Explanation

To find the temperature at which \( k_1 = k_2 \) for the given reactions, we can follow these steps:

### Step 1: Set the rate constants equal to each other
We have the two rate constants:
\[ k_1 = 10^{15} e^{-\frac{2000}{T}} \]
\[ k_2 = 10^{14} e^{-\frac{1000}{T}} \]
Setting them equal gives:
\[ 10^{15} e^{-\frac{2000}{T}} = 10^{14} e^{-\frac{1000}{T}} \]

### Step 2: Simplify the equation
We can divide both sides by \( 10^{14} \):
\[ 10 e^{-\frac{2000}{T}} = e^{-\frac{1000}{T}} \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation:
\[ 10 = e^{-\frac{1000}{T} + \frac{2000}{T}} \]
This simplifies to:
\[ 10 = e^{\frac{1000}{T}} \]

### Step 4: Take the natural logarithm of both sides
Taking the natural logarithm on both sides gives:
\[ \ln(10) = \frac{1000}{T} \]

### Step 5: Solve for T
Now we can solve for \( T \):
\[ T = \frac{1000}{\ln(10)} \]

### Step 6: Calculate the value of T
Using the value of \( \ln(10) \approx 2.303 \):
\[ T = \frac{1000}{2.303} \approx 434.2 \text{ K} \]

Thus, the temperature at which \( k_1 = k_2 \) is approximately **434.2 K**.

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