Chemistry 0620 · IGCSE · Rate of reaction

Rate of reaction — practice question

With manganese(IV) oxide acting as the catalyst, hydrogen peroxide, $\text{H}_2\text{O}_2$, breaks down to form water and oxygen. $2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)$
(a)[2]

What does the term catalyst mean?

(b(i))[1]

When is the rate of reaction greatest?

(b(ii))[1]

Suggest one way to increase the reaction rate while using the same amounts of hydrogen peroxide and manganese(IV) oxide.

(c(i))[1]

Calculate the amount of hydrogen peroxide, in moles, used in this experiment.

(c(ii))[1]

Use your answer to c(i) together with the equation to find the number of moles of oxygen formed in the reaction. $2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)$

(c(iii))[1]

Calculate the volume of oxygen produced at r.t.p.

(c(iv))[1]

What effect would increasing the mass of catalyst have on the volume of oxygen produced?

(c(v))[1]

Work out the volume of oxygen that would be formed if $20\,\text{cm}^3$ of $0.2\,\text{mol dm}^{-3}$ hydrogen peroxide were used instead of $20\,\text{cm}^3$ of $0.1\,\text{mol dm}^{-3}$ hydrogen peroxide.

(d)[3]

Describe how the second experiment is carried out. Make sure you state clearly how you would ensure that catalyst is the only variable.

Worked solution & mark scheme

This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: speeds up the reaction / increases the rate

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