Skip to main content
Ch.9 - Thermochemistry: Chemical Energy

Chapter 9, Problem 90

Citric acid has three dissociable hydrogens. When 5.00 mL of 0.64 M citric acid and 45.00 mL of 0.77 M NaOH are mixed at an initial temperature of 26.0 °C, the temperature rises to 27.9 °C as the citric acid is neutralized. The combined mixture ahs a mass of 51.6 g and a specific heat of 4.0 J/(g·°C). Assuming that no heat is transferred to the surroundings, cal- culate the enthalpy change for the reaction of 1.00 mol of cit- ric acid in kJ. Is the reaction exothermic or endothermic?

Verified Solution
Video duration:
6m
This video solution was recommended by our tutors as helpful for the problem above.
630
views
Was this helpful?

Video transcript

Hi everyone. This problem reads. There are three D. Sociable hydrogen and phosphoric acid, a mixture of 4.50 mL of 0.68 molar phosphoric acid and 57 mL of 570.53 molar potassium hydroxide has an initial temperature of 25 degrees Celsius. When the acid was neutralized, the final temperature of the mixture was 28.2 degrees Celsius. The total mass of the mc mr is 57.1 g. And it has a specific heat of four jewels per gram degrees Celsius. What is the entropy change for the reaction of one mole of phosphoric acid in killer jewels? If there is no heat transferred to the surroundings, is the reaction endo thermic or eggs? A thermic. So for this problem we have two things. We want to answer. The first one is what is the entropy change for the reaction? And the second is is the reaction endo thermic or eggs. A thermic. So let's go ahead and get started. So looking at our problem, the concentration of potassium hydroxide is an excess. So let's go ahead and calculate our moles of phosphoric acid. So we want to know what is our moles of phosphoric acid And we're gonna do that by starting off with our volume. So we're given a volume of 4.5 male leaders. Okay, so we want to go from volume to moles. Okay, so let's first start off by going from milliliters to liters. Okay. And one leader there is 1000 mL. So now we make sure our units cancel and we're no longer in units of milliliters. So now we want to go from leaders to moles and we can go from leaders to moles by using the mill arat E that was given. Okay, So we're told that the mill arat e. Is 10000.68 molar. And what that means is moles per liter. So when we write this out, well right there is 0.68 moles per liter. Okay, so with that our leaders cancel and we're left with units of moles. So let's go ahead and do this calculation. And when we do this calculation, we're going to get 0.00306 moles. Okay, so now that we know that we can calculate our heat of the reaction or we're going to calculate our heat of water. Okay, so we're told in the problem that the total mass of the mixture is 57.1 g. So we're going to take our 57 0.1 g and multiply it by the specific heat. The four jewels per gram degree Celsius. Okay, So we take our 51 g and multiply it by four jewels per degree or per gram Degree C. And we're going to multiply this by the temperature change. So our temperature changes going from 28.2°C to 25°C. So once we do this calculation, the heat for our water is equal to 730 .88 Jewels. So the heat of the reaction is going to equal the negative heat of water. So that makes the sign negative. So we have negative 730.88 jewels. So now we're going to calculate the entropy change for the reaction. The entropy change for the reaction is going to be Jules Permal. So we're going to take the heat of the reaction. So we have negative 730.88 jewels Permal. Okay. And the question asked for it and kill a jewel. So we need to convert this from Jules to kill a jules. So we're gonna write in one kill a jewel. There is 1000 jewels Jules cancel. And now we have killer joules per mole. So let's go ahead and do this calculation. And when we do it we get negative. 238.8497 kg jewels per mole of phosphoric acid. Okay, so that is going to be the entropy change for the reaction. And the second part of the question asked is the reaction endo thermic or eggs a thermic. So for this we need to look at our sign for our entropy change and here our sign is negative. So the reaction is going to be eggs a thermic. So let's go ahead and highlight our final answers. So our entropy change for the reaction is negative. 238.8497 kg jewels per mole. And the reaction is exo thermic. Okay, so that is it for this problem, I hope this was helpful.
Related Practice
Textbook Question
When 0.187 g of benzene, C6H6, is burned in a bomb calorimeter the temperature rises by 3.45 °C. If the heat capacity of the calorimeter is 2.46 kJ>°C, calculate the combustion energy 1∆E2 for benzene in units of kJ/g and kJ/mol.
1147
views
Textbook Question
When 1.50 g of magnesium metal is allowed to react with 200 mL of 6.00 M aqueous HCl, the temperature rises from 25.0 °C to 42.9 °C. Calculate ΔH in kilojoules for the reaction, assumign that the heat capacity of the calorimeter is 776 J/°C, that the specific heat of the final soltuion is the same as that of water [4.18 J(g·°C)] and that the density of the solution is 1.00 g/mL
769
views
Textbook Question
A 110.0 g piece of molybdenum metal is heated to 100.0 °C and placed in a calorimeter that contains 150.0 g of water at 24.6 °C. The system reaches equilibirum at a final temeprature of 28.0 °C. Calcualte the specific heat of molybdenum metal in J/g·°C. The specific heat of water is 4.18 J/g·°C
985
views
Textbook Question
What is Hess's law, and why does it 'work'?
596
views
Textbook Question
The following steps occur in the reaction of ethyl alcohol (CH3CH2OH) wiht oxygen to yield acetic acid (CH3CO2H). Show that equations 1 and 2 sum to give the net equation and calculate ΔH° for the net equation. (1) CH3CH2OH(l) + 1/2 O2(g) → CH3CHO (g) + H2O(l) ΔH° = -174.2 kJ (2) CH3CHO(g) + 1/2 O2(g) → CH3CO2H(l) ΔH° = -318.4 kJ (Net) CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) ΔH° = ?
682
views
Textbook Question
The industrial degreasing solvent methylene chloride, CH2Cl2, is prepared from methane by reaction with chlorine: CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g) Use the following data to calcualte ΔH° in kilojoules for the reaction: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔH° = -98.3 kJ CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g) ΔH° = -104 kJ
709
views