SO2 reacts with O2 as follows 2SO2+O2→ 2SO3.The rate of disappearance of SO2 is 2.4×10−4mol lit−1 min−1. Then

2SO2+O2→ 2so3Rate=−12dSO2dt=dO2dt=12dSO3dtRate=122.4×10−4=1.2×10−4dSO3dt=−12dSO2dt⇒dSO3dt=2.4×10−4⇒−12dSO2dt=dO2dt⇒122.4×10−4=dO2dt⇒dO2dt=1.2×10−4

SO2 reacts with O2 as follows 2SO2+O2→ 2SO3.The rate of disappearance of SO2 is 2.4×10−4mol lit−1 min−1. Then

2SO2+O2→ 2so3Rate=−12dSO2dt=dO2dt=12dSO3dtRate=122.4×10−4=1.2×10−4dSO3dt=−12dSO2dt⇒dSO3dt=2.4×10−4⇒−12dSO2dt=dO2dt⇒122.4×10−4=dO2dt⇒dO2dt=1.2×10−4

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$S O_{2}$ reacts with $O_{2}$ as follows $2 S O_{2} + O_{2} \overset{}{\to} 2 S O_{3} .$ The rate of disappearance of $S O_{2}$ is $2.4 \times 10^{- 4} m o l l i t^{- 1} \min^{- 1} .$ Then

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Rate of reaction is $1.2 \times 10^{- 4} m o l l i t^{- 1} \min^{- 1}$

Rate of appearance of $S O_{3}$ is $2.4 \times 10^{- 4} m o l l i t^{- 1} \min^{- 1}$

Rate of disappearance of $O_{2}$ is $1.2 \times 10^{- 4} m o l l i t^{- 1} \min^{- 1}$

Rate of reaction twice the rate of disappearances of $S O_{3}$

answer is A, B, C.

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Detailed Solution

$2 S O_{2} + O_{2} \overset{}{\to} 2 s o_{3}$
$\begin{array}{l} R a t e = \frac{- 1}{2} \frac{d [S O_{2}]}{d t} = \frac{d [O_{2}]}{d t} = \frac{1}{2} \frac{d [S O_{3}]}{d t} \\ R a t e = \frac{1}{2} (2.4 \times 10^{- 4}) = 1.2 \times 10^{- 4} \\ \frac{d [S O_{3}]}{d t} = \frac{- 1}{2} \frac{d [S O_{2}]}{d t} \Rightarrow \frac{d [S O_{3}]}{d t} = 2.4 \times 10^{- 4} \\ \Rightarrow \frac{- 1}{2} \frac{d [S O_{2}]}{d t} = \frac{d [O_{2}]}{d t} \\ \Rightarrow \frac{1}{2} (2.4 \times 10^{- 4}) = \frac{d [O_{2}]}{d t} \Rightarrow \frac{d [O_{2}]}{d t} = 1.2 \times 10^{- 4} \end{array}$