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Finding time t in a first order reaction (JEE Adv 2019)

In reference to the above question,taking into account the stoichiometric coefficient, the standard equation of first order reaction time changes by a factor of 2.

I came across a question which i find similar to the one above. However when i tried solving it the same way, i couldn't get the right answer. Can someone point out where did i go wrong?

The question is attached below:

The decomposition of $\ce{N2O5}$ in carbon tetrachloride was followed by measuring the volume of $\ce{O2}$ gas evolved: $$\ce{2 N2O5(solv) ->[CCl4] 2 N2O4(solv) + O2(g)}$$

The maximum volume of $\ce{O2}$ gas obtained was $\pu{100 cm3}$. In $\pu{500 min}$, $\pu{90 cm3}$ of $\ce{O2}$ were evolved. The first order rate constant (in $\pu{min-1}$) for the disappearance of $\ce{N2O5}$, is:

Here is what i did

$$t=\frac{1}{2k} \cdot \ln{\left(\frac{P_0}{P}\right)}$$

\begin{aligned} K &=\dfrac{2.303}{2×500} \log _{10}\left(\dfrac{200}{200-(2)90}\right) \\ =& \dfrac{2.303}{2×500} \log _{10}\left(\dfrac{200}{20}\right) \\ & \\ K &=\dfrac{2.303}{1000} \mathrm{~min}^{-1} \\ \end{aligned}

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  • $\begingroup$ It is not a good idea to use 0 instead of O in chemical formulas. $\endgroup$
    – Poutnik
    Commented Nov 14, 2023 at 6:26
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    $\begingroup$ You should derive the equation for k directly for the 1st order rate equation. And general advise - avoid using literal values too early. Do not use them when you do not have to yet. $\endgroup$
    – Poutnik
    Commented Nov 14, 2023 at 7:50
  • $\begingroup$ @Poutnik I have derived the equation and that is the reason why i am asking the error in the answer $\endgroup$
    – Kashvi
    Commented Nov 14, 2023 at 8:32
  • $\begingroup$ minor point - upper case $K$ is typically reserved for equilibrium constants (thermodynamics), while lower case $k$ is used for rate constants (kinetics). And no matter which you use, you should be consistent across the problem and not switch between them. $\endgroup$
    – Andrew
    Commented Nov 14, 2023 at 13:26

1 Answer 1

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As oxygen is reportedly being created by the first order kinetics, we can write:

$$- \frac{1}{2} \frac{\text{d}[\ce{N2O5}]}{dt} = r = k_\text{r}[\ce{N2O5}]$$

Let substitute $y = [\ce{N2O5}]$:

$$\frac{\text{d}y}{y} = -2k_\text{r}\cdot \text{d}t$$

$$\int_{y_0}^{y_1} {\frac{\text{d}y}{y}} = -2k_\text{r}\cdot \int_0^t {\text{d}t}$$

$$\ln {\frac{y_1}{y_0}} = -2k_\text{r}t$$

$$y_1 = y_0 \cdot \exp{(-2k_\text{r}t)}$$

$$y_0 - y_1 = y_0 (1 - \exp{(-2k_\text{r}t)} ) = \left( \frac{y_0}{V_\mathrm{\ce{O2}, fin}} \right) \cdot V_\ce{O2}$$


$$V_\ce{O2} = V_\mathrm{\ce{O2},fin} \cdot \left(1 - \exp {(-kt)} \right)$$ $$ \exp {(-kt)} = 1 - \frac{V_\ce{O2} }{ V_\mathrm{\ce{O2},fin}} $$ $$ k = \frac 1t \ln {\left(\frac{V_\ce{O2,fin} }{ V_\mathrm{\ce{O2},fin} - V_\mathrm{\ce{O2}}} \right)} $$

$$ k = \frac {1}{\pu{500 min}} \ln {\left(\frac{\pu{100 mL} }{ \pu{100 mL} - \pu{90 mL}} \right)} \approx \frac{2.303}{500 } \pu{min-1}$$

$$ k_\mathrm{r} = \frac{k}{2} = \frac{2.303}{1000 } \pu{min-1}$$


It is clear that $k = 2 k_\text{r}$, what meant the reaction kinetic rate constant is not the same as the first order rate constant, with the proportionality constant equal to stoichiometric coefficient. It is the direct consequence of how the reaction rate is defined.

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