The reaction N2O5(g) - 2NO2(g) + 1/2O2(g) is a first order reaction. The initial concentration of N

Q: Why do we use 2.303 in the formula?

The natural logarithm ($\ln$) is converted to base-10 logarithm ($\log_{10}$) by multiplying with 2.303. The formula $k = \frac{1}{t} \ln(C_0/C_t)$ becomes $k = \frac{2.303}{t} \log(C_0/C_t)$.

Q: Why did we convert 1 hour to 60 minutes?

The question specifically asks for the rate constant $k$ in the units of $\text{min}^{-1}$. If we used 1 hour, our answer would be in $\text{hr}^{-1}$.

Q: How did we calculate $\log 2$ when it wasn't given?

We used the property $\log 10 = 1$. Since $10 = 2 \times 5$, $\log 10 = \log 2 + \log 5$. Therefore, $\log 2 = 1 - \log 5$.

Q: Does the stoichiometry of the reaction affect the rate constant formula here?

No, for a simple first-order reaction given in terms of the reactant's concentration, the integrated rate law $k = \frac{2.303}{t} \log(C_0/C_t)$ applies directly to the reactant $N_2O_5$ regardless of the product coefficients.

Q: What if I didn't round off at the end?

The question explicitly asks for the answer as the 'Nearest integer' for the coefficient of $10^{-3}$. Entering a decimal like 6.755 would be marked incorrect in an integer-type question.

Question

The reaction $N_2O_5(g) \longrightarrow 2NO_2(g) + \frac{1}{2}O_2(g)$ is a first order reaction. The initial concentration of $N_2O_5$ is $2.40 	imes 10^{-2} 	ext{ mol L}^{-1}$ at $318 	ext{ K}$. The concentration of $N_2O_5$ after $1 	ext{ hour}$ was $1.60 	imes 10^{-2} 	ext{ mol L}^{-1}$. The rate constant of the reaction at $318 	ext{ K}$ is ...... $	imes 10^{-3} 	ext{ min}^{-1}$. (Nearest integer)
[Given : $\log 3 = 0.477, \log 5 = 0.699$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup
We are given a chemical reaction involving the decomposition of dinitrogen pentoxide ($N_2O_5$). The problem explicitly states that this is a **first-order reaction**.

We are provided with the initial concentration of the reactant, $C_0 = 2.40 	imes 10^{-2} 	ext{ mol L}^{-1}$, and its concentration after a certain time, $C_t = 1.60 	imes 10^{-2} 	ext{ mol L}^{-1}$. The time elapsed is $1 	ext{ hour}$.

Our goal is to find the rate constant $k$ in the units of $	ext{min}^{-1}$. Because the requested unit is per minute, our very first step must be to convert the given time from hours to minutes.
$$t = 1 	ext{ hour} = 60 	ext{ minutes}$$

### The Master Equation
For any first-order reaction, the integrated rate law relates the rate constant, time, and concentrations as follows:
$$k = \frac{2.303}{t} \log \left( \frac{C_0}{C_t} ight)$$

This equation is the heart of first-order kinetics. It tells us how the concentration of a reactant decays exponentially over time. Let's substitute our known values into this equation:
$$k = \frac{2.303}{60} \log \left( \frac{2.40 	imes 10^{-2}}{1.60 	imes 10^{-2}} ight)$$

### The Logarithm Trick
Notice how the $10^{-2}$ terms in the numerator and denominator perfectly cancel each other out. We are left with:
$$\frac{2.40}{1.60} = \frac{24}{16} = \frac{3}{2} = 1.5$$

So, our equation simplifies to:
$$k = \frac{2.303}{60} \log(1.5)$$

To evaluate $\log(1.5)$, we can use the properties of logarithms:
$$\log(1.5) = \log \left( \frac{3}{2} ight) = \log 3 - \log 2$$

The problem gives us $\log 3 = 0.477$ and $\log 5 = 0.699$. But wait, we need $\log 2$! Here is where we use a clever mathematical trick. We know that $\log 10 = 1$, and since $10 = 2 	imes 5$, we can write:
$$\log 2 = \log \left( \frac{10}{5} ight) = \log 10 - \log 5$$
$$\log 2 = 1 - 0.699 = 0.301$$

Now, we can easily find $\log(1.5)$:
$$\log(1.5) = 0.477 - 0.301 = 0.176$$

### Final Calculation
Let's plug this value back into our rate constant expression:
$$k = \frac{2.303}{60} 	imes 0.176$$
$$k = \frac{0.405328}{60} \approx 0.006755 	ext{ min}^{-1}$$

To match the format requested in the question ($\dots 	imes 10^{-3}$), we rewrite this as:
$$k = 6.755 	imes 10^{-3} 	ext{ min}^{-1}$$

The question asks for the **nearest integer**. Rounding off $6.755$, we get $7$.

Therefore, the value of the rate constant is $7 	imes 10^{-3} 	ext{ min}^{-1}$.

Enter Numerical Value:

Visualized Solution

\text{Given Data}

\text{First Order Rate Equation}

\text{Substituting the Values}

\text{Simplifying the Logarithm}

\text{Final Calculation}

The Sigma Insight: Rate of Chemical Reaction

Analyzing the Setup We are given a chemical reaction involving the decomposition of dinitrogen pentoxide ( $N_{2} O_{5}$ )

The Master Equation

The Logarithm Trick Notice how the $1 0^{- 2}$ terms in the numerator and denominator perfectly cancel each other out

Final Calculation

Similar Questions

For a first order reaction, $A ⟶ products$ the concentration of $A$ changes from $0.1 M$ to $0.025 M$ in $40 min$ . The rate of reaction when the concentration of $A$ is $0.01 M$ is

For a chemical reaction $A ⟶ B$ , it was found that concentration of $B$ is increased by $0.2 mol L^{- 1}$ in $30 min$ . The average rate of the reaction is $\dots \times 1 0^{- 1} mol L^{- 1} h^{- 1}$ . (Nearest integer)

Decomposition of $H_{2} O_{2}$ follows a first order reaction. In $50 min$ , the concentration of $H_{2} O_{2}$ decreases from $0.5$ to $0.125 M$ in one such decomposition. When the concentration of $H_{2} O_{2}$ reaches $0.05 M$ , the rate of formation of $O_{2}$ will be

For the following reaction $2 X + Y k P$ the rate of reaction is $\frac{d [ P ]}{d t} = k [X]$ . Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: $ln 2 = 0.693$ )

For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is ........... (Integer answer)

Rate constant, $k$ of the first order reaction when initial concentration $(C_{0})$ and concentration $(C_{t})$ at time $t$ , is given by equation $k t = lo g C_{0} - lo g C_{t}$ Graph is a straight line if we plot

For the reaction $2 H_{2} (g) + 2 NO (g) \to N_{2} (g) + 2 H_{2} O (g)$ the observed rate expression is, $rate = K_{f} [NO]^{2} [H_{2}]$ . The rate expression for the reverse reaction is

For a first-order reaction R $⟶$ P at a given temperature, $k$ is the rate constant. For this reaction, at the given temperature, the concentrations of R and P at a time $t$ are [R] and [P], respectively. The correct graphical representation(s) for this reaction is(are)