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Calorimeter
 

Calorimeter

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    Calorimeter Calorimeter Presentation Transcript

    • GROUP MEMBER’S:CHANG PEI XIAN D 20101037455CHAN SU FANG D 20101037450LECTURER :PN. FARIDAH YUSOFF
    • INTRODUCTION OF CALORIMETERA calorimeter (from Latin calor, meaning heat) is a device used for calculate the enthalpy change of reaction, the science of measuring the heat of chemical reactions or physical changes as well as heat capacity.The most common types of calorimeter is the bomb calorimeter(constant volume calorimeter) and coffee cup(constant pressure calorimeter).The other types include differential scanning calorimeters, isothermal micro calorimeters, titration calorimeters and accelerated rate calorimeters.
    • BASIC CONCEPT IN APPLICATION OF CALORIMETEREndothermic process occurs in which the system absorbs heat.Exothermic process occurs in which a system loses heat.The enthalpy change that accompanies a reaction is called the enthalpy of reaction, or merely the heat of reaction, and is sometimes written as ∆Hrxn, where rxn means reaction.
    • Since , ∆H = H final – H initial, the enthalpy change for a chemical reaction is given by the enthalpy of the products minus the enthalpy of the reactants: ∆H = H products – H reactantsTypically, we can determine the magnitude of the heat flow produces.The measurement of heat flow is calorimetry, and the device used to measure heat flow is a calorimeter.The heat capacity of an object is the amount of heat needed to raise the temperature by 1 K (or 1 C).
    • The heat capacity of one gram of a substance is called its specific heat capacity, or merely is called its specific heat.The specific heat, Cs, of a substance can be determined experimentally by measuring the temperature change, ∆T, that a known mass, m, of the substance undergoes when it gains or loses a specific quantity of heat, q: Specific heat = (quantity of heat transferred)(grams of substance) x (temperature change) Cs = q m x ∆T
    • BOMB CALORIMETER(CONSTANT-VOLUME CALORIMETER)
    • INTRODUCTION/THEORYCombustion reactions are most conveniently studied using a bomb calorimeter.Combustion is a process which an organic compound reacts completely with excess oxygen.Heat is released when combustion occurs. This heat is absorbed by the calorimeter contents, causing a rise in the temperature of the water.
    • To calculate the heat of combustion, we must know the total heat capacity of the calorimeter, Ccal. This quantity is determined by measuring the resulting temperature change.For example, the combustion of exactly 1 g of benzoic acid, C6H5COOH in a bomb calorimeter produces 26.38 kJ of heat. It increases the temperature by 4.857 0C. The heat capacity of the calorimeter is then given by Ccal = 26.38 kJ / 4.8570C = 5.341 kJ/0C. By knowing the value of Ccal , we can measure temperature changes produced by other reactions , and then we can calculate the heat evolved in the reaction, qrxn .
    • qrxn = Ccal x TSince the reactions in a bomb calorimeter are carried outunder constant-volume conditions, the heat transferredcorresponds to the change in internal energy, E, ratherthan the change in enthalpy, H .
    • EXAMPLE 1 Using Bomb Calorimeter Data to Determine a Heat of Reaction The combustion of 1.010 g sucrose, C12H22O11, in a bomb calorimeter causes the temperature to rise from 24.92 to 28.33 C. The heat capacity of the calorimeter assembly is 4.90 kJ/ C.a) What is the heat of combustion of sucrose expressed in kilojoules per mole of C12H22O11?b) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 Calories.
    • Solution:Calculate the qCalorimeter with the equationqCalorimeter = heat capacity of calorimeter x ∆ TqCalorimeter = 4.90 kJ/ C x (28.33 – 24.92) C = (4.90 x 3.41) kJ = 16.7 kJqreaction = - qCalorimeter = -16.7 kJThe heat of combustion of the 1.010 g sample is -16.7kJ.Per gram C12H22O11:qreaction = - 16.7 kJ/ 1.010 g C12H22O11 = -16.5 kJ/ g C12H22O11Per mole C12H22O11: qreaction = - 16.5 kJ/ g C12H22O11 x ( 342.3 g C12H22O11/ 1 mol C12H22O11) = - 5.65 x 103 kJ/ mol C12H22O11
    • COFFEE-CUP(CONSTANT-PRESSURE CALORIMETR)
    • INTRODUCTION/THEORYA very simple “ coffee-cup” calorimeter , is often used in general chemistry labs to illustrate the principles of calorimeter.This is because the calorimeter is not sealed, the reaction occurs under the essentially constant pressure of the atmosphere.
    • We can assume that the calorimeter itself does not absorb heat.For an exothermic reaction, heat is “lost” by the reaction and “gained” by the solution, so the temperature rises.For an endothermic reaction, heat is “lost” by the solution and “gained” by the reaction, so the temperature drops.The heat gained by the solution, qsoln equal to qrxn. qsoln = -qrxn qsoln = -qrxn = mc T
    • EXAMPLE 1Determining a Heat of Reaction from Calorimetric Data Two solutions, 25.00 mL of 2.50 M HCI(aq) and 25.00 mL of 2.50 M NaOH(aq), both initially at 21.1 C, are added to a Styrofoam-cup calorimeter and allowed to react. The temperature rises to 37.8 C. Determine the heat of the neutralization reaction, expressed per mole of H2O formed. Is the reaction endothermic or exothermic?
    • Solution:Because the reaction is a neutralization reaction, let us call the heat of reaction qneutralization. Now, according to equation qneutralization = - q calorimeterq calorimeter = 50.00 mL x (1.00 g/mL) x (4.18 J/g C) x (37.8 – 21.1) C = 3.5 x 103 Jq neutralization = - q calorimeter = - 3.5 x 103 J = - 3.5 kJ
    • In 25.00 mL of 2.50 M HCI, the amount of H+ is? mol H+ = 25.00 mL x (1 L/1000 mL) x (2.50 mol/ 1 L) x (1 mol H+/ 1mol HCI) = 0.0625 mol H+So , in 25.00 ml of 2.50 M NaOH there is 0.0625 mol of OH- .The H+ and the OH- combine to form 0.0625 mol H2OThe amount of heat produced per mole of H2O is q neutralization = ( - 3.5 kJ/0.0625 mol H2O) = - 56 kJ/ mol H2Ob)Because q neutralization is a negative value, the neutralization reaction is exothermic.