Questions: The temperature dependence of the reaction rate constant is given by the equation: ln k = ln A - (Ea) / (R T) Where: k = rate constant T = temperature A = frequency factor Ea = Energy of activation In order to solve for the Energy of activation, Ea, you must: Step One Add the same expression to each side of the equation to leave the term that includes the variable by itself on the right-hand side of the expression: + ln k = + ln A - (Ea) / (R T) Drag and drop your selection from the following list to complete the answer: ln (1/A) - ln A ln A

The temperature dependence of the reaction rate constant is given by the equation:
ln k = ln A - (Ea) / (R T)

Where:
k = rate constant
T = temperature
A = frequency factor
Ea = Energy of activation

In order to solve for the Energy of activation, Ea, you must:
Step One
Add the same expression to each side of the equation to leave the term that includes the variable by itself on the right-hand side of the expression:
+ ln k = + ln A - (Ea) / (R T)

Drag and drop your selection from the following list to complete the answer:
ln (1/A)
- ln A
ln A

Transcript text: The temperature dependence of the reaction rate constant is given by the equation: \[ \ln k=\ln A-\frac{E_{\mathrm{a}}}{R T} \] Where: \[ \begin{array}{l} k=\text { rate constant } \\ T=\text { temperature } \\ A=\text { frequency factor } \\ E_{\mathrm{a}}=\text { Energy of activation } \end{array} \] In order to solve for the Energy of activation, $E_{\mathrm{a}}$, you must: Step One Add the same expression to each side of the equation to leave the term that includes the variable by itself on the right-hand side of the expression: (Be sure that the answer field changes from light yellow to dark yellow before releasing your answer.) \[ +\ln k=\quad+\ln A-\frac{E_{\mathrm{a}}}{R T} \] Drag and drop your selection from the following list to complete the answer: $\ln \frac{1}{A}$ $-\ln A$ $\ln A$

Solution

Solution Steps

Step 1: Isolate the Term with $ E_{\mathrm{a}} $

To isolate the term that includes $ E_{\mathrm{a}} $ on the right-hand side, we need to eliminate $\ln A$ from the right-hand side. We do this by subtracting $\ln A$ from both sides of the equation:

\[ \ln k - \ln A = -\frac{E_{\mathrm{a}}}{R T} \]

Step 2: Simplify the Equation

The equation now becomes:

\[ \ln \left(\frac{k}{A}\right) = -\frac{E_{\mathrm{a}}}{R T} \]

This step simplifies the expression by combining the logarithms on the left-hand side.

Step 3: Solve for $ E_{\mathrm{a}} $

To solve for $ E_{\mathrm{a}} $, multiply both sides by $-R T$:

\[ E_{\mathrm{a}} = -R T \ln \left(\frac{k}{A}\right) \]

This step isolates $ E_{\mathrm{a}} $ on one side of the equation, providing the expression for the energy of activation.