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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Chemical Reaction and their Control
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[Post New]21 Feb 2008 23:20:24 IST
    Subject: Chemical Reaction and their Control
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shannyc (0)

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When 15.3g of sodium nitrate was added to 250.0 cm3 of water in a polystyrene cup calorimeter, the temperature FELL from 25.0 °C to 21.5 °C. Assume the specific heat capacity for the solution to be 4.18 J g-1  °C-1. Calculate the enthalpy change for 1 mole of the following reaction
 
NAaNO3 (s) + aq ----> Na+ (aq) + NO3 -(aq)
    
[Post New]21 Feb 2008 23:28:19 IST
    Subject: Chemical Reaction and their Control
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rockey (168)

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[Post New]21 Feb 2008 23:30:13 IST
    Subject: Re:Chemical Reaction and their Control
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chetan_kp (278)

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enthalpy change for 1 mole = amt. of heat liberated when 1 mole of the solute is dissolved in solvent

mass of water = 250 gm
fall in temp = 3.5 C
c = 4.18
heat liberated  = mcdt = 250*3.5* 4.18 = 3657.5 J

qty. of NaNO3 = 15.3 gm
Mol. mass of NaNO3 = 85
so no. of moles in 15.3 gm = 15.3/85 = 0.16moles
heat liberated for 0.16 moles = 3657.5 J
so heat liberated for 1 mole = 3657.5/0.16=  22.859 kJ
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