Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2 is 7.78 x 10-1 M/s, what is the rate of the loss of O3? 2 O3(g) → 3 O2(g)

The rate of appearance of oxygen is 0.52  M/s.

We have the equation of the reaction as;

2 O3(g) → 3 O2(g)

Rate of disappearance of O3 = -1/2d[O3]/dt

Rate of appearance of oxygen = 1/3d[O2]/dt

From the equations above, we can notice that;

d[O3]/dt = 2/3d[O2]/dt

We are told in the question that d[O2]/dt = 7.78 x 10-1 M/s

Hence;

d[O3]/dt = 2/3( 7.78 x 10-1 M/s)

d[O3]/dt = 0.52  M/s

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The rate of the loss of is 0.52M/s

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.

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 of disappearance of =

Rate of formation of =

Rate of formation of =

Thus Rate of disappearance of =

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

the answer is minerals.