Question

6. A student carries out a calorimetry experiment using HCI and NaoH. What effect will each of the following have on the calculated enthalpy of the reaction relative to the actual enthalpy of reaction? (larger, smaller or no change). Explain your reasoning. a. The thermometer always registered a temperature 0.30 oc higher than the actual temperature b. The calorimeter constant used in the calculations by the student is larger than the actual value. (Hint: The best way to answer this question is to take your calorimeter constant, add at least 5 or 10 to the value and quickly redo the calculations. After completing the calculations, you should be able to complete the following statement: As the calorimeter constant increases, the calculated enthalpy of the reaction c. Unfortunately, the HCI concentration was actually 1.90 M instead of the 2.00 M HCI used in the calculations. (Hint: do a similar calculation. This is an estimate of how errors affect the result. If your data is not matching the expected values, perhaps these questions will help you understand why.)

Answer 1

Ans. 6a. No change

While making calculation in calorimeter system, the change in temperature, that is, increase in temperature of the calorimeter system is noted. The change is temperature remains independent of an erroneously calibrated thermometer.   

For example,

Assume: actual initial temperature (A) = 25⁰C      ;erroneous initial temperature (B) = 25.3⁰C

            actual initial temperature (C) = 27⁰C       ;erroneous initial temperature (D) = 27.3⁰C

Now, difference in temperature (dT) = final – initial temperature

                        Case 1 (correctly reading thermometer) = C- A = 2⁰C

                        Case 2: (Incorrectly reading thermometer) = D- B = 2⁰C

Thus, dT remains the same in either case. It does not affect the result.

6b. Heat released by neutralization reaction is absorbed by calorimeter system to increase its temperature.

Suppose the total heat released during acid-base neutralization = 1000 J

So, the calorimeter system absorbs 1000 J.

Now using, q = m x c x dT                   

Where, q= heat change,                m= mass in gram

c= specific heat of calorimeter                                    dT = change in temperature

Using correct value of c,

q= 1000 J = mc1 dT

Using incorrect (higher) value of c, q =

q= 1000 J = mc2 dT

Where, C2 (incorrect value) > C1 (correct value)

                Say, C2 = 2 J/g⁰C               and C1= 1 J/g⁰C

Case 1 correct value adopted:                     dT = 1000 / m x 1 = (1000/ m)⁰C

Case 1 incorrect value adopted:                dT = 1000 / m x 2 = (500/ m)⁰C

Thus, incorrect (higher) value of specific heat of calorimeter gives smaller dT or smaller change in temperature. Smaller change in temperature i.e. little increase in temperature means that the reaction releases less heat. Thus, the calculated (incorrectly) value is less than actual value.

6c. Actual [HCl] = 1.9 M                 ; Incorrect [HCL] = 2.0 M

Suppose dT calculated = Y = 10                   [for time being all quantities are used without unit]

                                                c = 2

                                                m1 = 1.9 (correct)                            m2 = 2.0 (incorrect)

Correct value adopted:

                                                q1 = m x c x Y = 1.9 x 2x 10 = 38

Incorrect (higher, 2.0 M > 1.9 M) adopted:          

                                                q2 = m x c x Y = 2.0 x 2 x 10 = 40

Thus, you calculated that ‘enthalpy’ = 40                ; but in fact, it is 38

So, you get a higher value than an actual one.