Graph Interpretation Mr. Chapman Chemistry 30. Graph Reading is an Important Skill Particularly in chemistry, it is important for us to be able to look.

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Presentation transcript:

Graph Interpretation Mr. Chapman Chemistry 30

Graph Reading is an Important Skill Particularly in chemistry, it is important for us to be able to look at a graph and glean the necessary information from it. Sometimes this can be easy, but at other times it can be difficult to wrap your head around. We will go through some graphs and attempt to answer some questions.

Graph #1 This graph is one of the more simple ones that you will see. It is a plain concentration vs. time graph, and given its lack of numbers, only general questions about it can really be answered.

Graph #1 Questions 1.What is this a graph of? 2.What is this graph showing us about the concentrations of reactants and products? 3.At the beginning of the chemical reaction, what is present, reactants or products? 4.Are reactants or products favoured at equilibrium?

Graph #2 This graph is quite a bit more complicated, and it is something that we need to be able to interpret it and understand it.

Graph #2 Questions 1.What is happening in this graph? Are the reactants or products favoured at equilibrium? 2.What stress is applied to the equilibrium at time one? 3.How does the equilibrium react to this stress? What concentrations increase and decrease as a result?

Graph #2 Questions 4.After time two, are the reactants or the products favoured at equilibrium? 5.What does this mean must have happened to the equilibrium constant? 6.What is the only stress that could have been applied at time two that could result in this change?

Graph #3 For this graph, assume that the variables represent the following things: A – the forward reaction B – the reverse reaction C – the two reactions together

Graph #3 Questions 1.What is different about this graph as opposed to the other two previously seen? 2.Which rate is faster initially, the forward reaction or the reverse reaction? 3.What is established at c? How do you know this?