0 g sample of kbr dissolves in 300 ml of water at

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Unformatted text preview: kJ/ °C q = CCalorimiter × (T f - 292.03) K = 5.431 kJ / K × (T f - 292.03) K = 5.037 kJ T f = 292.96 K Molar Heat Capacity---The heat capacity of 1 mole of a mole substance. Unit : J/mol·K or J/mol·°C Molar Heat Capacity: Cmolar = q /(∆T × mole) q = Cmolar × ∆T × mole Specific Heat or Specific Heat Capacity --Specific The heat capacity of 1 g of a substance. Unit : J/g°C or J/g K J/g J/g Cs = q /(∆T × mass in grams) q = Cs × ∆T × mass in grams Question: Question: A 5.0 g sample of KBr dissolves in 30.0 mL of water at heat 25.0°C. The final temperature is 18.1°C. What is the heat of solution per gram of KBr? Specific heat of KBr = 0.435J/g°C. Specific heat of H2O = 4.184J/g°C Formation Formation of the solution: KBr(s) → K + (aq)+Br - (aq) 30g water 25.0°C → 18.1°C solution q total =q water +q KBr Heat is used as KBr dissolves: “endothermic” q = C m × ∆T × m ∆H > 0 ∆T = 18.1 -25.0 = -6.9 °C q water = 4.18 J/g°C × 6.9°C × 30.0 ml × 1.000g/ml = 866 J q KBr = 0.435 J/g°C × 6.9°C × 5.0 g= 15 J q total =q water +q KBr = 866 J/ + 15 J=881J The heat of solution per gram of KBr ∆H=q p = 881J ÷ 5.0 g=176.2 J/g (Endothermic) The heat of solution per mole of KBr q = 176.2 J/g × 119g/mol =2.097kJ/mol Enthalpies Enthalpies of Formation Standard enthalpy of formation (∆H o) : f The change in enthalpy for the reaction that form one mole of a compound from its most stable elements in their standard state most in 298.15K (1 atm and 298.15K). most standard All of most stable elements at standard state: ∆Hfo = 0 ∆H o (O 2(g) ) = 0 f ∆H o (O3 ) = 142kJ/mol f ∆H o (C, graphite) = 0 f ∆H o (C, diamond) = 1.90kJ/mol f ∆H (H 2(g) ) = 0 ∆H (H(g)) = 217.94 kJ/mole 0 f 0 f No way to measure the absolute value of the enthalpy of a substance Enthalpies Formation Enthalpies of Formation :∆Hf o ∆Hf o = standard heat of formation reaction standard Example: CH4 (g) is formed by this reaction: C(s, graphite)+2H 2 (g) → CH 4 (g) ∆H RXN = -74.8 kJ/mol at 25°C and 1 atm: ∆H °f of CH 4 ( g ) = - 74.8 kJ / mol Example: CO2 (g) is formed by this reaction: C(s, graphite)+O 2 (g) → CO 2 (g) ∆H RXN = -393.5 kJ/mol at 25°C...
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This document was uploaded on 02/26/2014.

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