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Teachers Notes: Problem 1:
Heat (q) = m x c x ΔΤ. ΔΤ = 76.0 °C - 4.0 °C = 72.0 °C.
Q = 300.0 g x 4.184 J/g ·°C x 72.0 °C = 90,400 J of
heat or 90.4 kJ of heat is released by the heater pad or gained
by the water.
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Problem 2:
Mg (s) + 2 H2O (l) → Mg(OH)2 (s) + H2 (g)
ΔHrxn = Σ ΔHproducts -ΣΔHreactants
ΔHrxn = (1 mol x -924.54 kJ/mol) - (2 mol x -285.830 kJ/mol)
= -352.88 kJ
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Problem 3:
Mg (s) + 2 H2O (l) → Mg(OH)2 (s) + H2 (g)
ΔSrxn = Σ Sproducts -Σ Sreactants
ΔSrxn = (63.18 J/K· mol +130.684 J/K· mol) -
(32.68 J/K· mol + 2 x 69.91 J/K· mol) =
21.364 J. The reaction in more disordered because the ΔSrxn
> 0
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Problem 4:
ΔGrxn = ΔHrxn - T ΔSrxn
ΔGrxn = -352.88 kJ - (298 K x 0.021364kJ) = -359.25 kJ
The reaction is spontaneous because the value for ΔGrxn <
0
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