Chemistry, asked by priyasingh25, 1 year ago

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

Answers

Answered by AkashMandal
66
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
Answered by skyfall63
21

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}

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