Question

Rate of formation of SO3 according to the reaction 2SO2+O2=2SO3 is 1.6×10 kg/min. Hence rate at which SO2 reacts is

Answer 1

rate of reaction= 1/2* Rate of disappearance of SO2 =1/2*Rate of formation of SO3

​=1/2* Rate of disappearance of SO2 = 1/2*1.6×10

therefore, Rate of disappearance of SO2=
2 × 8

Rate of disappearance of SO2= 16kg/min

Answer 2

The rate at which $\bold{SO_2}$ reacts is $\bold{0.8 \times 10^{-3} \ k g / \min}$

Explanation:

The given balanced reaction is as follows:

$2 S O_{2}+O_{2} \rightarrow 2 S O_{3}$

In terms, the rate of reaction represented is as follows:

$-\frac{1}{2} \frac{d\left[S O_{2}\right]}{d t}=-\frac{d\left[O_{2}\right]}{d t}=+\frac{1}{2} \frac{d\left[S O_{3}\right]}{d t}$

From given,

Rate of formation of $SO_3$ is $1.6 \times 10 \ \mathrm{kg} / \mathrm{min}$

Now we consider $O_2$ and $SO_3$

$-\frac{d\left[O_{2}\right]}{d t}=+\frac{1}{2} \frac{d\left[S O_{3}\right]}{d t}$

$=\frac{1}{2} \times 1.6 \times 10^{-3} \ \mathrm{kg} / \mathrm{min}$

$=0.8 \times 10^{-3} \ \mathrm{kg} / \mathrm{min}$

Therefore, rate at which $SO_2$ reacts is $0.8 \times 10^{-3} \ \mathrm{kg} / \mathrm{min}$