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Natali [406]
4 years ago
15

Suppose you have a calorimeter that contains 100.0 grams of water at an initial temperature of 25*C. A salt (2.19 g, 0.020 moles

) is dissolved in the water, and the final temperature is 29*C. Calculate the standard heat of solution (on a per mole basis). Was the dissolution exothermic or endothermic?
Chemistry
2 answers:
katen-ka-za [31]4 years ago
7 0
ΔH=MCΔT
ΔH=100 x 4.2 x 4
ΔH=1680

ΔH per mole = ΔH ÷ moles
ΔH per mole = 1680 ÷ 0.02
<span>ΔH per mole= 84000Jmol
</span>84000 ÷ 1000 = 84KJmol

its exothermic as heat is given out into the solution 
Tresset [83]4 years ago
4 0

<u>Answer:</u> The standard heat of the solution is 85512.5 J/mol and the reaction is endothermic in nature.

<u>Explanation:</u>

To calculate the heat released or absorbed, we use the equation:

q=mc\Delta T

where,

q = heat absorbed or released

m = mass of solution = (100 + 2.19) g = 102.19 g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = T_2-T_1=(29-25)^oC=4^oC

Putting values in above equation, we get:

q=102.19g\times 4.184J/g.^oC\times 4^oC\\\\q=1710.25J

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H^o_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = 1710.25 J

n = number of moles = 0.020 moles

\Delta H^o_{rxn} = standard enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{1710.25J}{0.020mol}=85512.5J/mol

  • Exothermic reactions are defined as the reaction in which heat is released and enthalpy change of the reaction is negative.
  • Endothermic reactions are defined as the reaction in which heat is absorbed and enthalpy change of the reaction is positive.

Hence, the standard heat of the solution is 85512.5 J/mol and the reaction is endothermic in nature.

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<u>Explanation</u>:

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(a)   ¹⁴₆ C ⇒         ¹⁴₇ N +  ⁰₋₁ e

(b) ¹⁴⁰₅₅ Cs ⇒  ¹⁴⁰₅₆ Ba + ⁰₋₁ e

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