Question

At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M-1s-1 : --->2SO3g+2SO2gO2g Suppose a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important.

Answer 1

Answer:

Approximately $0.245\; \rm M$. (Moles-per-liter.)

Explanation:

$\rm 2\; SO_3\; (g) \to 2\; SO_2\; (g) + O_2\; (g)$.

$\rm SO_3\; (g)$ is the only reactant in this reaction. Only the concentration of

Let $x = \left[\rm SO_3\; (g)\right]$ (that's the concentration of $\rm SO_3\; (g)$.) By "second-order" kinetics, the question likely means that the rate of change in $x$ (with respect to time $t$) is proportional to $x^2$. In other words,

$\displaystyle -\frac{dx}{dt} = k\, x^2$,

where $k$ is the rate constant of the reaction. Note the negative sign in front of the fraction. Reactants are consumed in a reaction, so their concentrations would become smaller.

Rearrange the equation to separate the variables:

$-\displaystyle \frac{1}{x^2}\, dx = k\, dt$.

Integrate both sides using the power rule for integration:

$\displaystyle \int -\frac{1}{x^2}\, dx = \int k\, dt$.

$\implies \displaystyle \frac{1}{x} = k\,t + C$.

The value of $C$ here is fixed; its exact value depends on the initial concentration of the reaction. Rearrange to obtain an equation for $x$ (concentration) with respect to $t$ (time.)

$x = \displaystyle \frac{1}{k\, t + C}$.

The concentration of $\rm SO_3\; (g)$ was $1.44\; \rm M$ at the beginning of the reaction. As a result, $C$ should ensure that $x = 1.44$ at $t = 0$.

Let $x = 1.44$, $t = 0$, and solve for $C$:

$\displaystyle 1.44 = \frac{1}{C}$.

$C \approx 0.694$.

According to the question, $k = 14.1$. Calculate the value of $x$ when $t = 0.240$:

$\begin{aligned}x &= \frac{1}{k\,t + C} \\ &= \frac{1}{14.1 \times 0.240 + 0.694} \\ &\approx 0.245\end{aligned}$.

Hence, the concentration of $\rm SO_3\; (g)$ is approximately $0.245\; \rm M$ after $0.240\; \rm s$.