For 2SO2+O2→2SO3, rate of disappearance of SO2is 4×10-3M-s-1 at t=10sec. Then, the amount of SO3 formed \& amount of O2 consumed at t=10sec respectively are

2SO2+O2⟶2SO3 −12d[SO2]dt=−d[O2]dt=12d[SO3]dt d[SO2]dt=4×10−3M/s −12[4×10−3]=−d[SO2]dt d[O2]dt=2×10−3M/s ln⁡1secd[O2]dt=2×10−3M ln⁡10secd[O2]dt=2×10−3×10M Molecular mass of O2=32g/mol Mass of O2 consumed=2×10−3×10×32=0.64g Similarly −12[4×10−3]=12d[SO3]dt d[SO3]dt=4×10−3M/s Molecular mass of SO3=80g/mol Mass of SO3 formed =4×10−3×10×80 =3.2gTherefore, the correct option is (D).

For 2SO2+O2→2SO3, rate of disappearance of SO2is 4×10-3M-s-1 at t=10sec. Then, the amount of SO3 formed \& amount of O2 consumed at t=10sec respectively are

2SO2+O2⟶2SO3 −12d[SO2]dt=−d[O2]dt=12d[SO3]dt d[SO2]dt=4×10−3M/s −12[4×10−3]=−d[SO2]dt d[O2]dt=2×10−3M/s ln⁡1secd[O2]dt=2×10−3M ln⁡10secd[O2]dt=2×10−3×10M Molecular mass of O2=32g/mol Mass of O2 consumed=2×10−3×10×32=0.64g Similarly −12[4×10−3]=12d[SO3]dt d[SO3]dt=4×10−3M/s Molecular mass of SO3=80g/mol Mass of SO3 formed =4×10−3×10×80 =3.2gTherefore, the correct option is (D).

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For $2 {SO}_{2} + O_{2} \to 2 {SO}_{3}$ , rate of disappearance of ${SO}_{2}$ is $4 \times 10^{- 3} M - s^{- 1}$ at $t = 10 \sec$ . Then, the amount of ${SO}_{3}$ formed \& amount of $O_{2}$ consumed at $t = 10 \sec$ respectively are

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$3.2 g; 0.64 g$

$0.1 g; 0.1 g$

$0.1 g; 0.2 g$

$0.016 g; 0.064 g$

answer is D.

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Detailed Solution

$\begin{aligned} 2 {SO}_{2} + O_{2} ⟶ 2 {SO}_{3} \\ - \frac{1}{2} \frac{d [{SO}_{2}]}{dt} = - \frac{d [O_{2}]}{dt} = \frac{1}{2} \frac{d [{SO}_{3}]}{dt} \\ \frac{d [{SO}_{2}]}{dt} = 4 \times 10^{- 3} M / s \\ - \frac{1}{2} [4 \times 10^{- 3}] = \frac{- d [{SO}_{2}]}{dt} \\ \frac{d [O_{2}]}{dt} = 2 \times 10^{- 3} M / s \end{aligned} \ln 1 \sec \frac{d [O_{2}]}{dt} = 2 \times 10^{- 3} M \ln 10 \sec \frac{d [O_{2}]}{dt} = 2 \times 10^{- 3} \times 10 M Molecular mass of O_{2} = 32 g / mol Mass of O_{2} consumed = 2 \times 10^{- 3} \times 10 \times 32 = 0 .64 g Similarly \begin{aligned} - \frac{1}{2} [4 \times 10^{- 3}] = \frac{1}{2} d \frac{[{SO}_{3}]}{dt} \\ \frac{d [{SO}_{3}]}{dt} = 4 \times 10^{- 3} M / s \end{aligned} Molecular mass of {SO}_{3} = 80 g / mol Mass of {SO}_{3} formed = 4 \times 10^{- 3} \times 10 \times 80 = 3.2 g$