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When 1.50 g of ba(s) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00°c to 33.10°c. if the specific heat of the solution is 4.18 j/(g ∙ °c), calculate δh for the reaction, as written. ba(s)+2h2o(l)→ba(oh)2(aq)+h2(g) δh=?

Respuesta :

Given:

Mass of Ba = 1.50 g

Mass of H2O = 100.0 g

Initial temp T1 = 22 C

Final Temp T2 = 33.1 C

specific heat c = 4.18 J/g c

To determine:

The reaction enthalpy

Explanation:

The heat released during the reaction is:

q = - mc(T2-T1) = - (100+1.5) g *4.18 J/g C * (33.1-22) C = -4709.4 J

# moles of Ba = Mass of Ba/Atomic mass of Ba = 1.5 g/137 g.mol-1 = 0.0109 moles

ΔH = q/mole = - 4709.4 J/0.0109 moles = - 432 kJ/mol

Ans : The enthalpy change for the reaction is -432 kJ/mol


pstnonsonjoku

The heat of reaction is -421 KJ/mol.

From the question, we have the following information;

mass of water =  100.00 g

Initial temperature = 22.00°c

Final temperature = 33.10°c

Heat capacity of water =  4.18 J/(g ∙ °c)

Number of moles of Ba =  1.50 g/137 g/mol = 0.0109 moles of Ba

The equation of the reaction is; Ba(s) + 2H2O(l) → Ba(OH)2(aq) + H2(g)

From; ΔH = mcθ

m =mass of water

c = heat capacity of water

θ = temperature difference

Substituting values;

ΔH = -(100.00 g ×  4.18 J/g°C × (33.10°c - 22.00°c)

ΔH = -4598 J

Note that ΔH is negative because heat was evolved.

For 0.0109 moles of Ba;

ΔH = -4598 J/0.0109 moles

ΔH = -421 KJ/mol

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