04-03-2018, 03:24 AM

Is my thinking correct? Have I used the correct equations? Thank you

Question:

You wish to find the enthalpy of reaction per mole of nitric acid for the following reaction:

Mg (s) + 2HNO3 (aq) --> Mg(NO3)2 (aq) + H2 (g).

At 25 degrees C, you add 0.050 grams of magnesium flakes to a makeshift coffee-cup calorimeter (Ccal= 11 Joules/C) that contains 100.0 mL of 1.25M HNO3. You close the top and monitor the temperature of the solution until it stabilizes at 27.25 degrees C. Assume the density of the solution is 1.01 g/mL and that the specific heat is 4.22 J/(g*°C).

a) What constitutes the system in this experiment?

My answer: Everything INSIDE the coffee-cup calorimeter.

b) WHat constitutes the surroundings in this experiment?

My answer: Everything OUTside the coffee-cup calorimeter

c) How are the heat of reaction and the heat of the surroundings related?

My answer: Heat gained or lost from the system is proportional to the heat gained or lost from the surroundings.

d) Calculate the heat gained or lost by the calorimeter?

My work:

(11 J/°C)*(27.25°C-25°C) = 24.75 joules?

e) Calculate the heat gained or lost by the solution?

My work:

(101 grams soln)*(4.184 J/g*°C)*(27.25°C-25°C) = 950.8 joules?

f) Calculate the heat gained or lost by the reaction?

My thoughts: I'm assuming because my answers to question a, b, and c, that energy gained/lost from the calorimeter is the energy gained or lost by the reaction. Is my thinking correct?

...If so, then the heat gained or lost by the reaction is equal to -24.75 joules

g) Calculate the enthalpy of reaction per mole of nitric acid?

My work:

(0.050 g Mg (solid))*(1 mol Mg/24.3050 g Mg)*(2 mol HNO3/1 mol Mg)= .411 moles HNO3 -> .9508 kJ/.411 moles = 2.31 kJ/mol HNO3

So that's it, if your wondering what I'm asking/looking for is simply confirmation that my thinking is correct and I'm doing the problems correctly. Thank you in advance.

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Question:

You wish to find the enthalpy of reaction per mole of nitric acid for the following reaction:

Mg (s) + 2HNO3 (aq) --> Mg(NO3)2 (aq) + H2 (g).

At 25 degrees C, you add 0.050 grams of magnesium flakes to a makeshift coffee-cup calorimeter (Ccal= 11 Joules/C) that contains 100.0 mL of 1.25M HNO3. You close the top and monitor the temperature of the solution until it stabilizes at 27.25 degrees C. Assume the density of the solution is 1.01 g/mL and that the specific heat is 4.22 J/(g*°C).

a) What constitutes the system in this experiment?

My answer: Everything INSIDE the coffee-cup calorimeter.

b) WHat constitutes the surroundings in this experiment?

My answer: Everything OUTside the coffee-cup calorimeter

c) How are the heat of reaction and the heat of the surroundings related?

My answer: Heat gained or lost from the system is proportional to the heat gained or lost from the surroundings.

d) Calculate the heat gained or lost by the calorimeter?

My work:

(11 J/°C)*(27.25°C-25°C) = 24.75 joules?

e) Calculate the heat gained or lost by the solution?

My work:

(101 grams soln)*(4.184 J/g*°C)*(27.25°C-25°C) = 950.8 joules?

f) Calculate the heat gained or lost by the reaction?

My thoughts: I'm assuming because my answers to question a, b, and c, that energy gained/lost from the calorimeter is the energy gained or lost by the reaction. Is my thinking correct?

...If so, then the heat gained or lost by the reaction is equal to -24.75 joules

g) Calculate the enthalpy of reaction per mole of nitric acid?

My work:

(0.050 g Mg (solid))*(1 mol Mg/24.3050 g Mg)*(2 mol HNO3/1 mol Mg)= .411 moles HNO3 -> .9508 kJ/.411 moles = 2.31 kJ/mol HNO3

So that's it, if your wondering what I'm asking/looking for is simply confirmation that my thinking is correct and I'm doing the problems correctly. Thank you in advance.

Logged