Question

Calculate the heat of sublimation for iodine if 25 g of iodine is heated from its solid form at -110°C to its gaseous form at 184°C. The molar heat capacity of iodine (s) is 54.44 J mol⁻¹ K⁻¹ and the molar heat capacity of iodine (g) is 36.57 J mol⁻¹ K⁻¹. The molar enthalpy of fusion of iodine is 15.52 kJ mol⁻¹.

Answer 1

To calculate the heat of sublimation for iodine, we need to consider the following steps:1. Heating solid iodine from -110C to its melting point  113.7C 2. Melting solid iodine to liquid iodine3. Heating liquid iodine from its melting point  113.7C  to its boiling point  184C 4. Vaporizing liquid iodine to gaseous iodineFirst, let's calculate the number of moles of iodine:Molar mass of iodine  I2  = 253.8 g/mol25 g / 253.8 g/mol  0.0985 molStep 1: Heating solid iodine from -110C to its melting point  113.7C q1 = n * C s  * Tq1 = 0.0985 mol * 54.44 J/mol*K *  113.7 -  -110  q1  1209.6 JStep 2: Melting solid iodine to liquid iodineq2 = n * Hfq2 = 0.0985 mol * 15.52 kJ/mol * 1000 J/kJq2  1529.6 JStep 3: Heating liquid iodine from its melting point  113.7C  to its boiling point  184C  can be skipped because the heat capacity of liquid iodine is not given. Step 4: Vaporizing liquid iodine to gaseous iodineWe need to find the molar enthalpy of vaporization  Hvap  first. We can use the following equation:Hsub = Hf + HvapWe know the molar enthalpy of sublimation  Hsub  for iodine is approximately 62 kJ/mol. So,Hvap = Hsub - HfHvap = 62 kJ/mol - 15.52 kJ/molHvap  46.48 kJ/molNow, we can calculate the heat required to vaporize liquid iodine:q4 = n * Hvapq4 = 0.0985 mol * 46.48 kJ/mol * 1000 J/kJq4  4574.2 JFinally, we can add up the heat required for all steps to find the total heat of sublimation:q_total = q1 + q2 + q4q_total = 1209.6 J + 1529.6 J + 4574.2 Jq_total  7313.4 JTherefore, the heat of sublimation for 25 g of iodine is approximately 7313.4 J.