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The rate of the following reaction is 0.180 M/s. What is the relative rate of change of each species in the reaction?

A+4B→2C
∆[A]/∆t=----------M/s
∆[B]/∆t=----------M/s
∆[C]/∆t=----------M/s

User Jolina
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1 Answer

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Final answer:

The rate of change for species A in the given reaction is -0.180 M/s, for species B it is -0.720 M/s, and for species C it is 0.360 M/s, based on the stoichiometry of the reaction.

Step-by-step explanation:

To find the relative rate of change of each species in the reaction A + 4B → 2C with a given overall rate of 0.180 M/s, we must consider the stoichiometry of the reaction. The change in concentration of A, ∆[A]/∆t, is equal to the negative of the reaction rate (0.180 M/s) because A is being consumed. Since A is being consumed at a rate of 0.180 M/s, B is being consumed four times as fast because the stoichiometry requires 4 moles of B for every mole of A. Therefore, the rate of change of B, ∆[B]/∆t, is -4 times the rate of change of A, which is -0.180 M/s multiplied by 4, yielding -0.720 M/s. Conversely, for C, the stoichiometry states that 2 moles of C are produced for every mole of A consumed, so the rate of change of C, ∆[C]/∆t, is 2 times the rate of change of A, resulting in a ∆[C] of 0.360 M/s.

Therefore, the rates of change for each species are:

  • ∆[A]/∆t = -0.180 M/s
  • ∆[B]/∆t = -0.720 M/s
  • ∆[C]/∆t = 0.360 M/s

User Kasongoyo
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