5. Determination of the product of the reaction - 2 points:

Total: 10 points.

## TASK 4

## SOLUTION:

To decrease the reaction rate, it is necessary to increase the volume of the system, i.e., to decrease the pressure and, thereby, reduce the concentration of the gaseous component - $\mathrm{NH}_{3}$. The concentration of $\mathrm{Br}_{2}$ will remain constant in this case.

The initial rate of the forward reaction was equal to:

$$
\mathrm{v} 1=\mathrm{k} \cdot\left[\mathrm{NH}_{3}\right]^{8} \cdot\left[\mathrm{Br}_{2}\right]
$$

When the concentration of ammonia increased, the rate of the forward reaction became

$$
\begin{gathered}
\text { equal to: } \\
v 2=\mathrm{k} \cdot\left[\mathrm{x} \cdot \mathrm{NH}_{3}\right]^{8} \cdot\left[\mathrm{Br}_{2}\right]=\mathrm{k} \cdot \mathrm{x}^{8} \cdot\left[\mathrm{NH}_{3}\right]^{8} \cdot\left[\mathrm{Br}_{2}\right] \\
\mathrm{v} 2 / \mathrm{v} 1=\mathrm{k} \cdot \mathrm{x}^{8} \cdot\left[\mathrm{NH}_{3}\right]^{8} \cdot\left[\mathrm{Br}_{2}\right] / \mathrm{k} \cdot\left[\mathrm{NH}_{3}\right]^{8} \cdot\left[\mathrm{Br}_{2}\right]=60
\end{gathered}
$$

After canceling all constants, we get

$$
\begin{gathered}
x^{8}=60 \\
x=1,66
\end{gathered}
$$

Thus, to decrease the reaction rate by 60 times, the volume must be increased by 1.66 times.
Answer: increase by 1.66 times.

## Scoring system: