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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 1</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 1</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 2</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 2</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 2</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 2</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 2</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
      <image:caption>Other than the two potential tricky scenario for calculating x values illustrated above, there is one last tricky situation may arise which is when you get two positive x values which are both are less than the initial concentration!  In that case, there is one further step you need to do in order to discard one of the two x values because you are only have one x value to determine equilibrium concentration of species. The extra step you need to perform is that you will need to substitute both x-values into the ICE table to determine the equilibrium concentration of the reacting species.  The x value that gives negative equilibrium concentration for one or more species in the ICE table will be the one that should be disregarded. This is because a substance cannot have a negative equilibrium concentration. Concentration (moles per litre) values must be positive. With that potential tricky scenarios cleared up, let’s continue in calculating our equilibrium concentration for all four of our species for our equilibrium reaction.</image:caption>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
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      <image:title>HSC Chemistry - Module 5 - Inquiry Question 3</image:title>
      <image:caption>In the ICE table: I stands for Initial or current concentration of the individual species participating in equilibrium reaction C stands for change in concentration of the individual species E stands for the equilibrium concentration of each individual species. NOTE: Sometimes, the question that you are given may provide you with ‘moles’ of the reacting species (reactants and/or products involving equilibrium reaction) rather than their concentrations. In that case, it may be easier for you to input their mole values rather than concentration into the ICE table. However, remember that you final goal of this question would be to determine the equilibrium concentration of each species. So, you will still need to convert the ‘moles at equilibrium’ of each species into ‘concentration at equilibrium’ after filling in the ICE table. However, in this question, the initial concentration of each species are provided in moles per litre (M) and so we don’t need to do anything fancy.  So, looking to the ICE table we have constructed for our question, the data have given us that the initial concentration of steam and carbon monoxide are both 0.500 moles per litre each. We put that in the table under their respective rows and columns. On the products side, we can assume that the initial concentration of hydrogen and carbon dioxide are zero as no values were given for them in the question. Notice that no data was given about the CHANGE in each of the four species’s concentration over time, i.e. for row labelled, ‘C’. However, we know that for equilibrium to be established, reactants’ concentration will decrease and the products will increase.  This was discussed last week in collision theory on equilibrium reactions about shifting equilibrium position in favour of the direction with greater number of successful collisions. For equilibrium to be established, there must be both reactants and products present. So, to establish equilibrium, steam will react with carbon monoxide to produce hydrogen and carbon dioxide. Since the molar ratio for all four species are the same, there are NO coefficients for ‘x’, i.e. the change in concentration.  P.S: Since x is the increase/decrease in some concentration number, the unit of x is in moles per litre (M). Hence, the equilibrium concentration for steam and carbon monoxide are their subtracting ‘x’ from their initial concentration as shown on the ICE table.  On the products side, the [H2] and [CO2] at equilibrium would be simply equal to x for both hydrogen and carbon dioxide. This is because they have an initial concentration of zero.  So, now we have the expression to determine the equilibrium concentrations for both the reactants and products. All we need to do now is to calculate the value of x!  If we can calculate x, we can then substitute the value of ‘x’ into the ICE table and determine the equilibrium concentration of each species involved in the equilibrium reaction!  In order to calculate x, there are some algebra calculations involved whereby the full calculation is shown below. Please Note: In the question, we were given that Keq (equilibrium constant) = 4.3.</image:caption>
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      <image:caption>NOTE: The halogen molecule (X2) that reacts with the unsaturated hydrocarbon (e.g. ethylene in this case) could be Cl2, F2, Br2, etc.</image:caption>
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