Decomposition of sodium Essay Example

Decomposition of sodium Essay In this experiment you are going to find ?Hreaction for the decomposition of sodium hydrogen carbonate:2NaHCO3 (s) Na2CO3(s) + CO2(g) +H2O(l) by an indirect method.You are going to experimentally determine ?Hreaction for the reactions:Na2CO3(s) + 2HCl(aq)2NaCl(aq) + CO2(g) +H2O(l)and NaHCO3(s)+ 2HCl(aq) NaCl(aq) + CO2(g) +H2O(l)and use your experimental values to find ?Hreaction for the decomposition of sodium hydrogen carbonate.In all calculations you may assume that the mass of any reaction mixture is the same as that of an equal volume of water (density of water=1.00g/cmà ¯Ã‚ ¿Ã‚ ½) and its specific heat capacity is the same as that of water (4180J/Kg/K).(i) To find ?Hreaction for Na2CO3(s) + 2HCl(aq)2NaCl(aq) + CO2(g) +H2O(l):? Weigh out accurately about 2.0g of anhydrous sodium carbonate.? Place 50cmà ¯Ã‚ ¿Ã‚ ½ of the hydrochloric acid provided (in excess) in a plastic beaker.? Record the temperature of the acid at half minute intervals. After 31/2 minutes add the anhydrous s odium carbonate, do not record the temperature.? Stir vigorously with the thermometer.? Record the temperature after 4 minutes (from the time you started recording) and continue to record the temperature of the mixture at 1/2 minute intervals up to 7 minutes (from the time you started recording).Record your results in a suitable format.Use your results to find ?Hreaction for Na2CO3(s) + 2HCl(aq)2NaCl(aq) + CO2(g) +H2O(l).READINGSTime Sodium Carbonate Sodium Hydrogen(mins) Carbonate0 19.4 19.41/2 19.5 19.41 19.5 19.511/2 19.6 19.52 19.5 19.521/2 19.5 19.53 19.6 19.631/2 4 22.0 16.441/2 22.0 16.15 22.0 16.151/2 22.0 16.26 21.9 16.261/2 21.9 16.37 21.9 16.5This table above shows the temperature change in à ¯Ã‚ ¿Ã‚ ½C.CALCULATIONSSodium CarbonateTemperature Rise from 19.4à ¯Ã‚ ¿Ã‚ ½C to 22.0à ¯Ã‚ ¿Ã‚ ½C (a rise of 2.6à ¯Ã‚ ¿Ã‚ ½C)=2.6KHeat energy given out = mass of mixture x specific heat capacity x temperature riseof mixtureM=0.05Kg Heat Capacity=4180J/Kg/K.Therefore heat given o ut= 0.05 x 4180 x 2.6 = 53.4 Kj.Using 2.07g of Na2CO3 (Mr=(232)+14+(163)=108)I.e. 2.07moles = 0.019167108?H=53.4/1000 = 2.786KJ/mol or 2.79KJ/mol0.019167