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A reacts, heterogeneously catalysed by X, to form B and C at temperature T according to the following balanced equation: $2A \xrightarrow{X} B + 3C$ The graph shows that 0.0024 moles per litre of A decomposes to produce 0.0036 moles per litre of C over 3000 seconds - Leaving Cert Chemistry - Question 9 - 2019

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Question 9

A-reacts,-heterogeneously-catalysed-by-X,-to-form-B-and-C-at-temperature-T-according-to-the-following-balanced-equation:--$2A-\xrightarrow{X}-B-+-3C$--The-graph-shows-that-0.0024-moles-per-litre-of-A-decomposes-to-produce-0.0036-moles-per-litre-of-C-over-3000-seconds-Leaving Cert Chemistry-Question 9-2019.png

A reacts, heterogeneously catalysed by X, to form B and C at temperature T according to the following balanced equation: $2A \xrightarrow{X} B + 3C$ The graph show... show full transcript

Worked Solution & Example Answer:A reacts, heterogeneously catalysed by X, to form B and C at temperature T according to the following balanced equation: $2A \xrightarrow{X} B + 3C$ The graph shows that 0.0024 moles per litre of A decomposes to produce 0.0036 moles per litre of C over 3000 seconds - Leaving Cert Chemistry - Question 9 - 2019

Step 1

Define rate of reaction.

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Answer

The rate of reaction is defined as the change in concentration of a reactant or product over time. Mathematically, it can be expressed as:

Rate=d[A]dt for reactants or d[B]dt for products\text{Rate} = -\frac{d[A]}{dt} \text{ for reactants or } \frac{d[B]}{dt} \text{ for products} where [A] is the concentration of reactant A.

Step 2

Use the graph to find the average rate of formation of C, in M s$^{-1}$. How does the instantaneous rate at 500 s compare with this average rate?

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Answer

The average rate of formation of C can be calculated using the graph data. The change in concentration of C over the entire time period is:

Average Rate=0.0036 moles/L3000 s=1.2×106 M s1\text{Average Rate} = \frac{0.0036 \text{ moles/L}}{3000 \text{ s}} = 1.2 \times 10^{-6} \text{ M s}^{-1}

At 500 s, the instantaneous rate of formation of C can be inferred from the slope of the tangent to the curve at that point, which is numerically smaller than the average rate.

Step 3

Explain the term heterogeneous catalysis.

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Answer

Heterogeneous catalysis refers to a process in which the catalyst is in a different phase from the reactants. This typically involves a solid catalyst and reactants in either gas or liquid form. The reaction occurs at the surface of the catalyst, where reactant molecules adsorb, react, and then desorb as products.

Step 4

Explain why pellets of X would be expected to be less effective as the catalyst in this reaction than the same mass of powdered X.

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Answer

Pellets of X have a smaller surface area compared to powdered X. The activity of a catalyst is largely dependent on its surface area, as more surface area allows for more active sites for reactions to occur. Therefore, powdered X will provide a greater number of active sites, translating to a higher reaction rate compared to pellets of the same mass.

Step 5

Without X this reaction has a large activation energy. Explain the underlined term. State and explain two ways of increasing the rate of this reaction, other than by using a catalyst.

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Answer

The activation energy is the minimum energy that reactant molecules must possess for the reaction to occur. It is the energy barrier that must be overcome for reactants to form products.

Two ways to increase the rate of reaction without a catalyst include:

  1. Increasing the temperature: Higher temperatures increase the kinetic energy of molecules, leading to more frequent and energetic collisions, thus increasing the rate of reaction.
  2. Increasing the concentration of reactants: This leads to more frequent collisions between reactant molecules, enhancing the likelihood of reactions occurring.

Step 6

Is this an equilibrium reaction? Explain your reasoning.

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Answer

Yes, the reaction can be considered an equilibrium reaction. This is because an equilibrium reaction is characterized by the forward and backward reactions occurring at the same rate, leading to constant concentrations of the reactants and products. In the context of this scenario, as A converts to B and C, the reverse reactions will eventually occur, stabilizing the concentrations over time.

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