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Chemistry Question on Chemical Kinetics

For the reaction, 2SO2+O2<=>2SO3,2SO_2 + O_2 <=> 2SO_3, the rate of disappearance of O2O_2 is 2×10−4 mol L−1 s−12 \times 10^{-4} \, mol \, L^{-1} \, s^{-1} . The rate of of appearance of SO3SO_3 is

A

2×10−4 mol L−1 s−12 \times10^{-4}\, mol\, L^{-1}\,s^{-1}

B

4×10−4 mol L−1 s−14 \times10^{-4} \, mol\, L^{-1}\,s^{-1}

C

1×10−4 mol L−1 s−11 \times10^{-4} \,mol\, L^{-1}\,s^{-1}

D

6×10−4 mol L−1 s−16 \times10^{-4} \,mol\, L^{-1}\,s^{-1}

Answer

4×10−4 mol L−1 s−14 \times10^{-4} \, mol\, L^{-1}\,s^{-1}

Explanation

Solution

Δ[O2]Δf=2×10−4\frac{\Delta [O_2]}{\Delta f} = 2 \times 10^{-4}
[SO3]Δt=2Δ[O2]Δt\frac{[SO_3]}{\Delta t} = 2 \frac{\Delta [O_2]}{\Delta t}
=2×2×10−4= 2 \times 2 \times 10^{-4}
=4×10−4 mol l−1s−1= 4 \times 10^{-4} \,mol \, l^{-1} s^{-1}