## Midterm 3B

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Siddiq 1E
Posts: 106
Joined: Fri Aug 09, 2019 12:15 am

### Midterm 3B

Can someone explain this question from the midterm?
Which combination of solutions of HCl and NaOH would produce the largest ∆T? Ans _____B
(A) 50 mL of 1 M HCl with 50 mL of 1 M NaOH
(B) 50 mL of 2 M HCl with 50 mL of 2 M NaOH
(C) 100 mL of 1 M HCl with 50 mL of 2 M NaOH
(D) 100 mL of 1 M HCl with 100 mL of 1 M NaOH

jisulee1C
Posts: 149
Joined: Thu Jul 25, 2019 12:17 am

### Re: Midterm 3B

Because the two solutions HCl (a strong acid) and NaOH (a strong base) are reacting, the most enthalpy is produced because the answer choice B has both in their highest concentrations (both at 2M). Both strong will create the largest delta T in their strongest concentration because a neutralisation reaction is exothermic (it releases heat). When all the base has been neutralised there is no reaction on the addition of more acid, and no more heat is released.

Ryan Yee 1J
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

### Re: Midterm 3B

The logic that I used in selecting B was that answer B had the highest concentration of both acid and base, and the least amount of water. While the B and D had the same number of moles, B had the least amount of water which means that it could change the temperature the greatest.

### Who is online

Users browsing this forum: No registered users and 1 guest