Sulphur dioxide has a melting point of -73 C and boiling point of -10C. What is

Question

Sulphur dioxide has a melting point of -73 C and boiling point of -10C. What is the heat required to raise the temperature of solid 6 Kg SO2 from melting point to 60C. dHfus = 8.619 KJ/mole and dHvap = 25.73 KJ/mol. Specific heat of liquit Sl = 0.995 J/gK and of gas is 0.622 J/g.K Here is what I did. # Moles of SO2 = 6000/64 = 93.75 moles Heat to convert from Solid to liquid Q1 = dHfus * Moles = 8.619 * 93.75 = 808.03 KJ Heat to raise the temp of liquid from -73 to -10 Q2 = m*Sl*(-10 - (-73) ) = 6000*0.995* 63/1000 = 376.11 KJ Heat to convert from liquid to Gas Q3 = dHvap * Moles = 25.73 * 93.75 = 2412.19 KJ Heat to raise the temp of Gas to 60 C Q4 = m *Sg*(60 - (-10) ) = 6000 * 0.622 * 70/1000 = 261.24 KJ Total heat consumerd to raise the temp of 6 Kg of SO2 from melting point to 60 C is = Q1 + Q2 + Q3 + Q4 = 808.03 + 376.11 + 2412.19 + 261.24 = 3857.57 KJ. What am I doing wrong here?

Explanation / Answer

I don't find anything wrong in the solution..its correct.

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