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Problems are on first law and enthalpy with answers
Typology: Exercises
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2Cl(g)® Cl2(g) ∆H° = – -‐‑243.4 kJ
(g)
(s)
(s)
(s)
(s)
(s)
(s)
® Na
3
Bomb calorimetery
Hesses Law
Heats of formation
the product. When the reaction occurs, heat is released, stabilizing the system and the
system moves to a lower energy potential.
2(l)
2(g)
2(g)
(l)
2(l)
2(g)
2(g)
(l)
[2] The enthalpy of reaction is ∆H° prod - ∆H° react. The reactant is at a higher potential
than the product. When the reaction occurs, heat is released, stabilizing the system and the
system moves to a lower energy potential.
[3] The reaction is exothermic. The ∆H is negative.
2.4 g MgX
1 mol Mg
24.0 g Mg
= 0.10 mol Mg X
2 mol Mg
= – 60.2 kJ
96 kJ X
2 mol MgO
40.0 g MgO
1 mol MgO
= 6.4 g MgO
7.50 g MgO X
1 mol MgO
40.0 g MgO
+1204 kJ
2 mol MgO
= 113 Kj
[4] q =
g–K
g Cu X (88.5°C – 25.0°C) X
[5] First, find the energy that the solution either absorbed or released. Then relate that
amount of energy to the moles of NaOH
q sol = (100.0 g water + 6.50 g NaOH) X (37.8°C - 21.6°C) X
g°C
6.50 g
39.90 g NaOH
1 mol NaOH
= 44.3 kJ/mol
B ® C 60 kJ B 30 kJ 90 kJ
Total 90 kJ C 60 kJ
[7] Hess’s law is a law of summations. Partial paths add to the desired path
Given the following enthalpies of reaction:
6(s)
4(s)
2(g)
∆H = +1640.1 kJ
4(s)
2(g)
10(s)
∆H = - 2940.1 kJ
6(s)
2(g)
10(s)
2(g)
2(g)
(g)
∆H = - 537 kJ
(s)
2(g)
4(g) ∆H = - 680 kJ 2
4(g)
(s)
2(g)
∆H = – 52.3 kJ
4(g)
2(g)
4(g)
(g)
∆H = – 1949.3 kJ