Question

Consider the following reaction: N204----- > 2NO2 Based on the above reaction a graph of ln k vs 1/T is plotted as shown below. (i) Calculate the activation energy for this reaction. (use R = 8.31 Jmol-1K -1 ) (ii) List two cases in which the rate of a reaction is equal to the rate constant. (Note: This is not specific to only the reaction above.)

Answer 1

(i) The activation energy for this reaction is $41.55 \text{ kJ mol}^{-1}$.

(ii) Two cases in which the rate of a reaction is equal to the rate constant are:

1. When the reaction is a zero-order reaction.
2. When the concentration of all reactants is unity (1 M or 1 atm).

Answer 2

The problem consists of two parts: calculating the activation energy from a given graph and identifying cases where the reaction rate equals the rate constant.

(i) Calculate the activation energy for this reaction.

The relationship between the rate constant ($k$), activation energy ($E_a$), and temperature ($T$) is given by the Arrhenius equation:

$k = A e^{-E_a/RT}$

Taking the natural logarithm on both sides, we get:

$\ln k = \ln A - \frac{E_a}{RT}$

This equation can be rearranged to match the form of a straight line, $y = mx + c$:

$\ln k = \left(-\frac{E_a}{R}\right) \left(\frac{1}{T}\right) + \ln A$

Here, $y = \ln k$, $x = \frac{1}{T}$, the slope $m = -\frac{E_a}{R}$, and the y-intercept $c = \ln A$.

From the given graph of $\ln k$ versus $1/T$, we can determine the slope of the line. Let's pick two distinct points from the line shown on the graph:

Point 1: $(1/T_1, \ln k_1) = (0.0028 \text{ K}^{-1}, 0.0)$ Point 2: $(1/T_2, \ln k_2) = (0.0032 \text{ K}^{-1}, -2.0)$

The slope ($m$) is calculated as:

$m = \frac{\Delta(\ln k)}{\Delta(1/T)} = \frac{\ln k_2 - \ln k_1}{(1/T_2) - (1/T_1)}$

$m = \frac{-2.0 - 0.0}{0.0032 \text{ K}^{-1} - 0.0028 \text{ K}^{-1}}$

$m = \frac{-2.0}{0.0004 \text{ K}^{-1}}$

$m = -5000 \text{ K}$

Now, we equate the calculated slope to $-\frac{E_a}{R}$:

$-\frac{E_a}{R} = -5000 \text{ K}$

$E_a = 5000 \text{ K} \times R$

Given the gas constant $R = 8.31 \text{ J mol}^{-1}\text{ K}^{-1}$:

$E_a = 5000 \text{ K} \times 8.31 \text{ J mol}^{-1}\text{ K}^{-1}$

$E_a = 41550 \text{ J mol}^{-1}$

To express activation energy in kJ mol$^{-1}$:

$E_a = \frac{41550}{1000} \text{ kJ mol}^{-1}$

$E_a = 41.55 \text{ kJ mol}^{-1}$

(ii) List two cases in which the rate of a reaction is equal to the rate constant.

The rate law for a general reaction is expressed as:

Rate = $k[A]^x[B]^y...$

where $k$ is the rate constant, $[A]$, $[B]$ are the concentrations of reactants, and $x$, $y$ are the orders of reaction with respect to those reactants.

For the rate of a reaction to be equal to the rate constant (Rate = $k$), the product of the concentration terms raised to their respective orders must be equal to unity, i.e., $[A]^x[B]^y... = 1$.

Two cases satisfy this condition:

1. For a zero-order reaction:

   In a zero-order reaction, the rate of reaction is independent of the concentration of the reactants. This means the order with respect to all reactants is zero.

   Rate = $k[A]^0[B]^0...$

   Since any non-zero number raised to the power of zero is 1, we have:

   Rate = $k \times 1 \times 1...$

   Rate = $k$

2. When the concentration of each reactant is unity (1 M or 1 atm):

   If the concentration of each reactant involved in the rate law is 1 M (for solutions) or 1 atm (for gases), then:

   Rate = $k[1]^x[1]^y...$

   Rate = $k \times 1 \times 1...$

   Rate = $k$

   This condition holds true regardless of the order of the reaction (e.g., first order, second order).