TBR general chem question

[pizza1994]

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Can someone please please look at page 175 question 45 in thermochemistry. I have no clue what is going on. This is about Mg(s) into Beaker 1.

Im extremely confused about what is happening 🙁


Thanks for all help 🙂

 

[sillyjoe]

Can someone please please look at page 175 question 45 in thermochemistry. I have no clue what is going on. This is about Mg(s) into Beaker 1.

Im extremely confused about what is happening 🙁


Thanks for all help 🙂

This passage describes an experiment where you obtain the delta H (enthalpy) for a reaction by allowing it to occur in an insulated system to minimize heat loss to the environment which would bring about error to the calculation. In this experiment they are measuring the enthalpy of the oxidation of Mg --> Mg 2+ and the reduction of 2 protons by putting it in an insulated beaker and measuring the temperature at regular time intervals.

The plot that they made represents the temperature vs time after the addition of the reactants to beaker 1. You can then find delta T and use the equation q=m*C*deltaT to obtain the enthalpy of the reaction. Now what is important to note for this experimental procedure is that by the time the thermometer reaches the temperature of the solution, some of the heat has been lost to the surroundings and the delta H calculated would be off. Therefore, you can extrapolate the line to t = 0 to get the initial delta T which is ~38.5-39 degrees.

I would suggest you review pages 152 to 154 and make sure you understand those concepts. Once you do this passage is not too difficult.

 

[pizza1994]

This passage describes an experiment where you obtain the delta H (enthalpy) for a reaction by allowing it to occur in an insulated system to minimize heat loss to the environment which would bring about error to the calculation. In this experiment they are measuring the enthalpy of the oxidation of Mg --> Mg 2+ and the reduction of 2 protons by putting it in an insulated beaker and measuring the temperature at regular time intervals.

The plot that they made represents the temperature vs time after the addition of the reactants to beaker 1. You can then find delta T and use the equation q=m*C*deltaT to obtain the enthalpy of the reaction. Now what is important to note for this experimental procedure is that by the time the thermometer reaches the temperature of the solution, some of the heat has been lost to the surroundings and the delta H calculated would be off. Therefore, you can extrapolate the line to t = 0 to get the initial delta T which is ~38.5-39 degrees.

I would suggest you review pages 152 to 154 and make sure you understand those concepts. Once you do this passage is not too difficult.

no im still confused because the inital tempertaure isnt it 22 as stated in the passage? and then how is final temperature 38? I dont get how you looked at that graph and said final temeprature is 38....because i see that 38 is the maximum temp but the it decreases and it seems that final temp is abut 32.

so for q=mcT

I know that m= 1.21 grams and c= 4.34 and T= UNKOWN and then also what is q?

 

[pizza1994]

no im still confused because the inital tempertaure isnt it 22 as stated in the passage? and then how is final temperature 38? I dont get how you looked at that graph and said final temeprature is 38....because i see that 38 is the maximum temp but the it decreases and it seems that final temp is abut 32.

so for q=mcT

I know that m= 1.21 grams and c= 4.34 and T= UNKOWN and then also what is q?

also just to say the question just in case: "adding 1.21 grams of Mg(s) into Beaker 1 instead of 2.43 grams would have led to a temperature increase up to what temperature?"

 

[sillyjoe]

also just to say the question just in case: "adding 1.21 grams of Mg(s) into Beaker 1 instead of 2.43 grams would have led to a temperature increase up to what temperature?"

First, think of the experiment. They are measuring heat change in water. The initial temperature was 22 degrees but it wasn't plotted on the graph because they started to use the thermometer after the experiment began. There first data point was therefore around 34 degrees because the reaction already started.

Thermometers take time to heat up, so there is a lag in between the time it is getting an accurate recording and the time of the peak temperature of the reaction. This brings error into the calculations because you do not have the peak temperature. Note that the slope of this graph is negative. This indicates heat is being lost to the surroundings, which in an ideal situation would not happen because you are measuring the total enthalpy of the reaction in a closed system. So our data is off. How do we solve this? We extrapolate a line back as indicated in the graph below. This solves both problems. We can assume that the slope is fairly constant, so if we extrapolate back we can get the temperature that is a very close approximation to the maximum temperature increase brought about by this reaction.

As far as your other question goes, if you cut the mass of the reactant in half, you cut the heat released (q) from the reaction by half. It is that simple.

 

[pizza1994]

First, think of the experiment. They are measuring heat change in water. The initial temperature was 22 degrees but it wasn't plotted on the graph because they started to use the thermometer after the experiment began. There first data point was therefore around 34 degrees because the reaction already started.

Thermometers take time to heat up, so there is a lag in between the time it is getting an accurate recording and the time of the peak temperature of the reaction. This brings error into the calculations because you do not have the peak temperature. Note that the slope of this graph is negative. This indicates heat is being lost to the surroundings, which in an ideal situation would not happen because you are measuring the total enthalpy of the reaction in a closed system. So our data is off. How do we solve this? We extrapolate a line back as indicated in the graph below. This solves both problems. We can assume that the slope is fairly constant, so if we extrapolate back we can get the temperature that is a very close approximation to the maximum temperature increase brought about by this reaction.

As far as your other question goes, if you cut the mass of the reactant in half, you cut the heat released (q) from the reaction by half. It is that simple.

holy crap I wont even lie but I really appreciate your initative to help me understand this!!!! THANKS super duper much!!! wow.....now the line extrapolation concept makes sense. appreciate it 🙂