The graph in Fig. 19.1 shows the volume of oxygen produced when hydrogen peroxide solution decomposes, with and without a catalyst. Fig. 19.1 (a) State whic...
Question 1 Report
The graph in Fig. 19.1 shows the volume of oxygen produced when hydrogen peroxide solution decomposes, with and without a catalyst.
Fig. 19.1
(a) State which curve, P or Q, shows the reaction with the catalyst. [1]
(b) Explain how the graph shows that the catalyst increases the rate of reaction. [2]
(c) Both curves reach the same final volume of gas. Explain why. [2]
(d) Use the graph to find how much longer the reaction without the catalyst takes to produce 30 cm³ of oxygen. [2]
Answer Details
This question tests how a catalyst affects a reaction, read from a volume-of-oxygen against time graph for the decomposition \( 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \). A catalyst speeds the reaction but does not change how much product forms.
(a) Curve P shows the reaction with the catalyst [1], because it rises far more quickly.
(b) The graph shows the catalyst increases the rate because curve P is steeper and reaches its final volume in a shorter time than curve Q [1], which means the catalysed reaction is faster [1].
(c) Both curves reach the same final volume because the catalyst does not change the amount of product [1]; the same amount of hydrogen peroxide decomposes, so the same volume of oxygen is made [1]. The catalyst provides a lower-energy pathway but is not used up and does not add extra reactant.
(d) Reading across at 30 cm3: curve P reaches 30 cm3 at about 5 s and curve Q at about 24 s [1], so the uncatalysed reaction takes about \( 24 - 5 = 19\ \text{s} \) longer (accept 16 to 22 s) [1].
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