How to do 4A.13

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Kavya Immadisetty 2B
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How to do 4A.13

Postby Kavya Immadisetty 2B » Tue Jan 28, 2020 6:02 pm

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter (q 5 23.50 kJ), resulting in a temperature rise of 7.32 8C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 8C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

Can someone explain how to do this question thanks!

Anuradha S 1F
Posts: 50
Joined: Fri Sep 27, 2019 12:29 am

Re: How to do 4A.13

Postby Anuradha S 1F » Tue Jan 28, 2020 6:44 pm

we have to use the equation qcal=CΔT.

so for the first reaction, we can solve for C by plugging in qcal (-3.50 kJ) and ΔT (7.32°C). We then get that C = -0.478 kJ/°C.

then for the second reaction, we use this C value to solve for qcal. Plug in the C value we solved above (-0.478 kJ/°C) and ΔT of the second reaction (2.49°C). We get that qcal for the second reaction = -1.19kJ.

Now we can solve the change in internal energy also known ΔU. ΔU = qreact+w and since there is no work done because there is no expansion, ΔU = qreact. And qreact = -qcal therefore qreact = 1.19kJ and therefore ΔU=1.19kJ


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