What is the rate of reaction (M/s) for the following process if the rate of change for NOBr was measured as -0.5 M/s.

2NOBr(g) ↔ 2NO(g) + Br2(g)

Answers

Answer 1

The rate of the reaction for the following process if the rate of change for NOBr was measured as -0.5 M/s. is 0.25 M/s.

The rate of a reaction is typically expressed as the rate of change of the concentration of a reactant or product over time. For the given reaction:

[tex]2NOBr(g)[/tex] ↔[tex]2NO(g)[/tex] [tex]+[/tex] [tex]Br_{2}[/tex](g)

The rate of the reaction can be expressed as:

[tex]rate = -1/2 * d[NOBr]/dt[/tex]

where [tex]d[NOBr]/dt[/tex] is the change in the concentration of NOBr over time, and the negative sign indicates that the concentration of NOBr is decreasing over time.

If the rate of change for NOBr was measured as -0.5 M/s, we can substitute this value into the above equation to find the rate of the reaction:

rate = -1/2 * (-0.5 M/s)

= 0.25 M/s

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