Question

Thinking about the previous problem: "A known reaction has AU = 38 kJ/mol. If the reaction is done and you measure the amount of work done to be w +2 kJ/mol, what is q in kJ/mol?" If you do the same reaction at constant volume, what is q in kJ/mol?

Answer 1

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PART 1:

Given that, $\Delta U=38\;kJ/mol$

Work done, $W=+2\;kJ/mol$

According to First law of thermodynamics, Heat energy is given by,

$q=\Delta U+W=38\;kJ/mol+2\;kJ/mol$

$\boldsymbol{\therefore q=40\;kJ/mol}$

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PART 2:

We know that, Work done is given by, $W=P\Delta V$

Where, $\Delta V$ is change in volume.

If the reaction happened at constant volume then, change in volume $\Delta V$ will be zero. Hence, work done will be zero.

Therefore, According to First law of thermodynamics, Heat energy is given by,

$q=\Delta U+W=38\;kJ/mol+0\;kJ/mol$

$\boldsymbol{\therefore q=38\;kJ/mol}$

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