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saul85 [17]
3 years ago
8

The reaction of solid calcium carbonate with hydrochloric acid is a heterogeneous reaction (a solid with a liquid). The rate law

for this reaction will have the following form: rate= k[HCL]n. Explain why the concentration of calcium carbonate does not appear in the rate law.
Chemistry
2 answers:
Umnica [9.8K]3 years ago
7 0
The concentration of calcium carbonate does not appear in the rate law because it is a solid and its concentration is constant. The molar density, which is fixed, is the concentration of the solid. In this chemical reaction, only the amount of hydrochloric acid affects the rate of reaction.
aleksandrvk [35]3 years ago
5 0

Answer:

Rate of reaction does not depend on concentration of reactants that are constant.

Explanation:

The rate law or rate equation for a chemical reaction is an equation that links the reaction rate with the concentrations or pressures of the reactants and constant parameters (normally rate coefficients and partial reaction orders).

Sincethe rate law pretty much shows the variation between the rate of reaction at different concentrations, it follows that a fixed concentration would have a fixed rate and as such rate of reaction is independendent on such reactant.

This is alswo what is happening here, in the reaction between solid calcium carbonate with hydrochloric acid, the reaction is independent on the concentration hence it is not represented in the rate law. The reason for this is because the concentrationof the solid calcium carbonate is constant and does not change. Same thing also happens in reactions involving large amount of water.

In short, rate of reaction does not depend on concentration of reactants that are constant.

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The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval
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<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t  

The average rate of disappearance of ozone ... is found to  

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm  

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rate = -(1/2)∆[O3]/∆t  

rate = -(1/2)*(-7.25 × 10^–3 atm)

      = 3.625 × 10^–3 atm  

Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

3.625 × 10–3 atm = +(1/3)∆[O2)/t  

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

              = 1.0875 x 10-2 atm over the same time interval

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