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Question 1:

A compound A reacts according to the following hypothetical equation and has a molecular weight of 48.36 g/mol.

3 A (s) +B (aq) ––> 2 C (aq) ∆H° = ?

A sample of A, weighing 0.152 g reacts in a flask containing 250.00 g of water and the water temperature increases from 24.85 °C to a temperature of 26.26 °C. Calculate ∆H° for the reaction as written in the equation.

Answer: -1408 kJ

Question 2:

For which of the following reactions is ∆H° = ∆H°f, the heat of formation?

i. C (s) + 2 F2(g) ––> CF4 (g) ∆H° = – 221.0 kJ ii. H(g) + Br (g) ––> HBr (g) ∆H° = –366.2 kJ iii. 2 C(s) + H2(g) + 3 Cl2(g) ––> 2 CHCl3 (g) ∆H° = –268.2

Answer: i only

Also, could someone explain exactly what delta H means in terms of Thermochemistry. I would like to have a better understanding of it.

Thank you very much! :)

For the first question I attempted to use the q=m x s x (delta) T for the water and I attempted to use stochiometry for substance A but I always receive the wrong answer. Could someone help me on where to start with these types of problems?

Water calculation:

(250) x (4.184) x (26.26-24.85)

= 1474.86 kJ

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H is enthalpy.

By definition H = U + PV (where U is internal energy, P is pressure and V is volume).

For question 1, your work says:

(250) x (4.184) x (26.26-24.85)

= 1474.86 kJ

Because you failed to show units in your work, you are off by a factor of 1000.

But the main reason your answer is different from the desired answer is you have calculated the amount of heat released by 0.152 g reacting, whereas the desired answer comes from calculating the heat released by 3 moles of A reacting, considering change in enthalpy of the system to be negative when heat is released to the surroundings.

For question 2:

Answer ii is wrong because the monoatomic forms of hydrogen and bromine are not the standard states of the elements. Answer iii is wrong because the equation is written for 2 moles being formed rather than 1 mole.

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  • $\begingroup$ So I calculated the heat released by .152 g of the substance which is -1474.86 J. But how would I get from 1474 J to 1408 kJ? $\endgroup$ – dsmith Dec 9 '14 at 18:04
  • $\begingroup$ Oh, I understand know. I caculated what would be .152 g of heat released. This would be .003 moles of substance a. The question asked for three moles of substance a. So I took 1474.86 and divided it by .00314 moles to get a ratio for substance a. I got 469700 J I then multiplied it by 3 which gave me 1409101.911. Then divide the answer by 1000, I get 1409 kJ, an approximate answer. Is this the correct method to use. I want to ensure that my logic is sound and that I didn't arrive to the correct answer from luck. Essentially it takes a release of 1409 kJ of energy for 3 moles of a to interact w $\endgroup$ – dsmith Dec 9 '14 at 18:12
  • $\begingroup$ @dsmith First, find how many moles of A equal 0.152 g. Then calculate how much heat would be released if 3 moles reacted. $\endgroup$ – DavePhD Dec 9 '14 at 18:14
  • $\begingroup$ Almost right. You just need to understand why there is a negative sign in the answer. If the temperature of the water increases, the change in enthalpy of the reaction is negative. $\endgroup$ – DavePhD Dec 9 '14 at 18:20
  • $\begingroup$ Yes the negative sign indicates that it is an exoteric reaction. When the water temperature increased it absorbed 1474 J. That means that .152 g of a released -1474.86. This would make all my caculations negative. $\endgroup$ – dsmith Dec 9 '14 at 18:30

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