Question

The initial concentration of N₂O₅ in the given first order reaction:

N₂O₅(g)→2NO₂(g)+$\frac{1}{2}$O₂(g)

was 1.20 × 10⁻² mol L⁻¹ at 318K. The concentration of N₂O₅ after 46 minutes was 0.20 × 10⁻² mol L⁻¹. The value of rate constant of the reaction at 318 K is x min⁻¹. Then find the value of 1000 X.

(Given log 3 = 0.48, log 2 = 0.30 ) (In 10 = 2.3)

Answer 1

39

Answer 2

The reaction is N₂O₅(g) → 2NO₂(g) + ½O₂(g). The reaction is first order with respect to N₂O₅. The integrated rate law for a first-order reaction is given by:

$k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right)$

where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time $t$, and $k$ is the rate constant.

Given:

• Initial concentration of N₂O₅, $[A]_0 = 1.20 \times 10^{-2}$ mol L⁻¹
• Concentration of N₂O₅ after $t = 46$ minutes, $[A]_t = 0.20 \times 10^{-2}$ mol L⁻¹

Substitute these values into the integrated rate law:

$k = \frac{1}{46 \text{ min}} \ln\left(\frac{1.20 \times 10^{-2} \text{ mol L⁻¹}}{0.20 \times 10^{-2} \text{ mol L⁻¹}}\right)$ $k = \frac{1}{46} \ln\left(\frac{1.20}{0.20}\right)$ $k = \frac{1}{46} \ln(6)$

We are given log 3 = 0.48, log 2 = 0.30, and ln 10 = 2.3. We know that $\ln(y) = \log_{10}(y) \times \ln(10)$. Using the given values:

$\log_{10}(6) = \log_{10}(2 \times 3) = \log_{10}(2) + \log_{10}(3) = 0.30 + 0.48 = 0.78$.

Now, convert $\log_{10}(6)$ to $\ln(6)$ using the given $\ln(10) = 2.3$:

$\ln(6) = \log_{10}(6) \times \ln(10)$ $\ln(6) = 0.78 \times 2.3$

Calculate $0.78 \times 2.3$: $0.78 \times 2.3 = 1.794$

So, $\ln(6) = 1.794$.

Now substitute the value of $\ln(6)$ back into the equation for $k$:

$k = \frac{1}{46} \times 1.794$ min⁻¹ $k = \frac{1.794}{46}$ min⁻¹

Calculate the value of $k$:

$k = 0.039$ min⁻¹

The value of the rate constant is given as $x$ min⁻¹. So, $x = 0.039$.

We need to find the value of $1000x$.

$1000x = 1000 \times 0.039$ $1000x = 39$