Calculate the amount of heat(in KJ) Camphor (MM = 152.2 g/mol; 0.990 g. mI^-1) i

Question

Calculate the amount of heat(in KJ) Camphor (MM = 152.2 g/mol; 0.990 g. mI^-1) is used as a solvent in various high temperature chemical measurements; it melts at 175 degree C and boils at 208 degree C. Given that you have a 50.0 ml, sample of pure camphor at 199.5 degree C, calculate the amount of heat (in kj) one must remove to lower this sample's temperature to 10 degrees below its melting temperature. Some physical and thermochemical data for camphor are as follows(not all data may be needed): Delte H = 6.820 kj mol, Delta H = 41.64 kj/mol; Cmp (s) = 271.2 J . mol^-3 K^-4 ; Cm.p (1) = 348.00 j . mol^-1 K^-1; Cm.p(vapor) = 167.34 j . mol^-1 K^-1.

Explanation / Answer

step 1 : cool camphor from 199.5 oC to 175 oC

q1 = mCpdT = (50 x 0.99/152.2) x 348 x (175-199.5) = -2.773 kJ

step 2 : convert liquid camphor to solid at 175 oC

q2 = mdH = (50 x 0.99/152.2) x -6.82 = -2.22 kJ

step 3 : cool solid from 175 oC to 165 oC

q3 = mCpdT = (50 x 0.99/152.2) x 271.2 x (165-175) = -0.882 kJ

So the total amount of heat to be removed for the process,

q = q1 + q2 + q3 = -5.875 kJ

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