28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of the solution by 6.9 degrees C. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer.

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28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of the
solution by 6.9 degrees C. How much would the temperature have increased if 100 mL of acid and 100 mL of base
had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer.
Transcribed Image Text:28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of the solution by 6.9 degrees C. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer.
For context, here is Example 5.5:
Example 5.5
Heat Produced by an Exothermic Reaction
When 50.0 mL of 1.00 M HCI(aq) and 50.0 mL of 1.00 M NAOH(aq), both at 22.0 °C, are added to a coffee
cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate
amount of heat produced by this reaction?
HСC(ag) + NaOН(ад) —
NaCl(aq) + H2O(1)
Solution
To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction
took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have
100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution
temperature reaches 28.9 °C.
The heat given off by the reaction is equal to that taken in by the solution. Therefore:
9 reaction =-9 solution
(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat
from the reaction, and there is no heat exchange between the calorimeter and the outside environment.)
Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature
change:
9 solution = (c Xmx AT)olution
To proceed with this calculation, we need to make a few more reasonable assumptions or approximations.
Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass
values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 x 102 g (two
significant figures). The specific heat of water is approximately 4.184 J/g °C, so we use that for the specific
heat of the solution. Substituting these values gives:
9 solution = (4.184 J/g °C)(1.0x 10 g)(28.9 °C- 22.0 °C) = 2.9 x 10 J
Finally, since we are trying to find the heat of the reaction, we have:
9reaction - solution =-2.9x 10 J
The negative sign indicates that the reaction is exothermic. It produces 2.9 kJ of heat.
Check Your Learning
When 100 mL of 0.200 M NaCl(ag) and 100 mL of 0.200 M AGNO3(aq), both at 21.9°C, are mixed in
a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is
produced by this precipitation reaction? What assumptions did you make to determine your value?
Answer: 1.34 x 10 J; assume no heat is absorbed by the calorimeter, no heat is exchanged between the
calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for
water
Transcribed Image Text:For context, here is Example 5.5: Example 5.5 Heat Produced by an Exothermic Reaction When 50.0 mL of 1.00 M HCI(aq) and 50.0 mL of 1.00 M NAOH(aq), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction? HСC(ag) + NaOН(ад) — NaCl(aq) + H2O(1) Solution To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C. The heat given off by the reaction is equal to that taken in by the solution. Therefore: 9 reaction =-9 solution (It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and the outside environment.) Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change: 9 solution = (c Xmx AT)olution To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 x 102 g (two significant figures). The specific heat of water is approximately 4.184 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives: 9 solution = (4.184 J/g °C)(1.0x 10 g)(28.9 °C- 22.0 °C) = 2.9 x 10 J Finally, since we are trying to find the heat of the reaction, we have: 9reaction - solution =-2.9x 10 J The negative sign indicates that the reaction is exothermic. It produces 2.9 kJ of heat. Check Your Learning When 100 mL of 0.200 M NaCl(ag) and 100 mL of 0.200 M AGNO3(aq), both at 21.9°C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value? Answer: 1.34 x 10 J; assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water
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