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Step-by-Step Solution
Step 1: Write down the expressions for k₁ and k₂
We are given:
$k_{1} = 10^{16} \cdot e^{-\frac{2000}{T}}$
$k_{2} = 10^{15} \cdot e^{-\frac{1000}{T}}$
Step 2: Set k₁ equal to k₂
We need to find the temperature $T$ at which $k_{1} = k_{2}$. Therefore,
$10^{16} \cdot e^{-\frac{2000}{T}} = 10^{15} \cdot e^{-\frac{1000}{T}}$
Step 3: Simplify the equation
Divide both sides by $10^{15}$:
$10^{16} / 10^{15} = 10^{1} = 10$
Thus, we get:
$10 \cdot e^{-\frac{2000}{T}} = e^{-\frac{1000}{T}}$
Step 4: Take the natural logarithm or common logarithm on both sides
Taking the common (base 10) logarithm on both sides for convenience, we have:
$\log \left( 10 \cdot e^{-\frac{2000}{T}} \right) = \log \left( e^{-\frac{1000}{T}} \right)$
Recall the logarithm properties:
$\log(ab) = \log(a) + \log(b)$
$\log(e^{x}) = \frac{x}{2.303}$ (approximation when log is base 10)
Hence,
$\log(10) + \log\left(e^{-\frac{2000}{T}}\right) = \log\left(e^{-\frac{1000}{T}}\right)$
$\log(10) = 1$, and
$\log\left(e^{-\frac{2000}{T}}\right) = -\frac{2000}{T} \cdot \frac{1}{2.303}$
$\log\left(e^{-\frac{1000}{T}}\right) = -\frac{1000}{T} \cdot \frac{1}{2.303}$
Step 5: Substitute and solve for T
Putting these into the equation, we get:
$1 - \frac{2000}{2.303\, T} = -\frac{1000}{2.303\, T}$
Rearrange terms:
$1 = -\frac{1000}{2.303\, T} + \frac{2000}{2.303\, T}$
$1 = \frac{2000 - 1000}{2.303\, T} = \frac{1000}{2.303\, T}$
Multiply both sides by $2.303\, T$:
$2.303\, T = 1000$
Finally,
$T = \frac{1000}{2.303}\, K$
Step 6: State the conclusion
Therefore, the temperature at which $k_{1} = k_{2}$ is
$\displaystyle \frac{1000}{2.303}\,\mathrm{K}.$
