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Consider the reaction “2 SO2 (g) + O2 (g) = 2 SO3 was 0.175 M. After 50 s the concentration of SO2 Date: (g)”. Initial concentration of SO2 (g) (g) became 0.0500 M. Calculate rate of the reaction

User Kdau
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1 Answer

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22 votes

Answer:

The answer is "
1.25 * 10^(-3) \ (m)/(s)"

Step-by-step explanation:

Calculating the rate of the equation:


=-(1)/(2) (\Delta [SO_2])/(\Delta t) =-(\Delta [O_2])/(\Delta t)= +(1)/(2) (\Delta [SO_3])/(\Delta t)\\\\=(\Delta [SO_2])/(\Delta t)=(0.0500-0.175\ M)/(505)= -2.5 * 10^(-3) \ (m)/(s)\\\\

Rate:


=(-2.5 * 10^(-3))/(2)=1.25 * 10^(-3) \ (m)/(s)

User RatajS
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