Question

The following equation is obtained for a first order reaction at 300 K.
$\log_{10} \frac{k}{A} = 0.00174$ What is the activation energy (in J mol\(^{-1}\)) of the reaction ?
(R = 8.314 J mol\(^{-1}\) K\(^{-1}\))

• A. 10.0
• B. 100.0
• C. 0.1
• D. 1.0

Hint:

The Arrhenius equation in its logarithmic form is $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$. This can be rearranged to $\log_{10} \left(\frac{k}{A}\right) = - \frac{E_a}{2.303 RT}$. When solving for activation energy ($E_a$), pay close attention to the signs. Activation energy ($E_a$) is typically a positive value. If the $\log_{10} (k/A)$ term is positive, it might indicate that the magnitude is intended, and the negative sign from the formula should be applied for calculation to get a positive $E_a$.

Answer 1

The Correct Option is A

Step 1: Recall the Arrhenius equation and its logarithmic form.
The Arrhenius equation describes the temperature dependence of reaction rates:
$k = A e^{-E_a / RT}$
where:
$k$ = rate constant
$A$ = pre-exponential factor (Arrhenius factor)
$E_a$ = activation energy
$R$ = gas constant
$T$ = absolute temperature
Taking the natural logarithm ($\ln$) on both sides: $\ln k = \ln A - \frac{E_a}{RT}$ Converting to base-10 logarithm ($\log_{10}$), we use the relationship $\ln x = 2.303 \log_{10} x$:
\ $2.303 \log_{10} k = 2.303 \log_{10} A - \frac{E_a}{RT}$ Dividing by 2.303:
$\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$ Rearranging the equation to match the given form:
$\log_{10} k - \log_{10} A = - \frac{E_a}{2.303 RT}$
$\log_{10} \left(\frac{k}{A}\right) = - \frac{E_a}{2.303 RT}$ 
Step 2: Substitute the given values into the equation.
Given:
$\log_{10} \frac{k}{A} = 0.00174$
Temperature, $T = 300\operatorname{K}$ Gas constant, $R = 8.314\operatorname{J mol^{-1} K^{-1}}$ Substitute these values into the derived equation:
$0.00174 = - \frac{E_a}{2.303 \times 8.314\operatorname{J mol^{-1} K^{-1}} \times 300\operatorname{K}}$ 
Step 3: Solve for the activation energy ($E_a$).
First, calculate the denominator:
$2.303 \times 8.314 \times 300 \approx 5744.142$ Now, rearrange the equation to solve for $E_a$:
$E_a = - 0.00174 \times (2.303 \times 8.314 \times 300)$
$E_a = - 0.00174 \times 5744.142$
$E_a \approx - 10.0$
There seems to be a sign discrepancy in the given question if the expected answer is positive. The $\log_{10} (k/A)$ term is typically negative because $k<A$ for a reaction with positive activation energy. 
If $\log_{10} (k/A)$ is given as positive, it implies either a negative activation energy (which is physically unusual for most reactions) or a misprint in the sign of the given value. 
Assuming the formula $\log_{10} \left(\frac{k}{A}\right) = - \frac{E_a}{2.303 RT}$ is correct and $E_a$ is positive, then $\log_{10} (k/A)$ should be negative.
If the question implies that the magnitude is $0.00174$ and we are looking for a positive $E_a$, then we use:
$| \log_{10} \left(\frac{k}{A}\right) | = \frac{E_a}{2.303 RT}$ $0.00174 = \frac{E_a}{5744.142}$ $E_a = 0.00174 \times 5744.142$ $E_a \approx 10.0$ Given that the options are positive and the most common scenario is a positive activation energy, it is highly likely that the value $0.00174$ in the problem statement should have been $-0.00174$. 
Assuming this is a common type of exam question where the absolute value is implicitly considered for calculation, we proceed with the positive result. The activation energy $E_a$ is approximately $10.0\operatorname{J mol^{-1}}$. 
The final answer is $\boxed{10.0}}$.