Question

a reaction 2x—>y+3z is being carried out in a closed vessel. the rate of disappearance of x,-∆[x]/ ∆t is found to be 0.066mol l-1s-1.calculate ∆[z]/∆t

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Answer 1

$+\frac{\Delta [z]}{dt}=0.099molL^{-1}s^{-1}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2x\rightarrow y+3z$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{\Delta [x]}{2dt}=+\frac{\Delta [y]}{dt}=+\frac{\Delta [z]}{3dt}$

Given: $-\frac{\Delta [x]}{dt}]=0.066molL^{-1}s^{-1}$

As : $-\frac{\Delta [x]}{2dt}=+\frac{\Delta [z]}{3dt}$

$+\frac{\Delta [z]}{dt}=\frac{3}{2}\times -\frac{\Delta [x]}{dt}=\frac{3}{2}\times 0.066molL^{-1}s^{-1}=0.099molL^{-1}s^{-1}$

Thus $+\frac{\Delta [z]}{dt}=0.099molL^{-1}s^{-1}$

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