The entropy change for the conversion of 1 mole of `alpha-" tin"` (at `13^(@)C, 1` atm) to 1 mole of `beta-" tin "(13^(@)C, 1 atm)` if the enthalpy of transition is `2.095 kJ mol^(-1)`

The entropy change of the conversion of 1 mol of alpha - tin (at 13^(@)C , 1 atm) to 1 mol of beta - tin( 13^(@) , 1 atm). If enthalpy of transition is 2.095 kJ mol^(-1) is :

What is the entropy that takes place in conversion of 1 mol of alpha- tin to 1 mol beta - tin at 17^(@)C , 1 atm pressure if enthalpy of transition is 2.90 kJ mol^(-1) ?

Calculate the enthalpy change for the following reversible process: alpha = T "in" hArr beta - T "in at" 13^(@)C 1 mol at atm 1 mol at 1 atm (Delta H_("trans") = 2288 J mol^(-1))

Calculate the entropy change accompanying the following change of state 5 mol of O_(2) (27^(@)C, 1 atm) rarr 5 mol of O_(2) (117^(@)C, 5 atm) C_(P) for O_(2) = 5.95 cal deg^(-1) mol^(-1)

Calculate the entropy change accompanying the following change of state 5 mol of O_(2) (27^(@)C, 1 atm) rarr 5 mol of O_(2) (117^(@)C, 5 atm) C_(P) for O_(2) = 5.95 cal deg^(-1) mol^(-1)

Calculate the entropy change accompanying the conversion of 1 mole of ice at 273.1 K and 1 atm pressure into steam at 373.1 K and 1 atm pressure. At 273.1 K, the molar heat of fusion of ice, DeltaH_(f) is 6.00 kJ mol^(-1) and at 373.1 K, the molar heat of vapourization of water, DeltaH_(v) , is 40.6 kJ mol^(-1) . Also assume that the molar heat capacities, C_(p) , in the temperature range 373.1 to 273.1 K remains constant. Given that C_(p) = 75.25 m mol^(-1)K^(-1) and log 13.66 = 1.1354.

Calculate the entropy change accompanying the conversion of 1 mole of ice at 273.1 K and 1 atm pressure into steam at 373.1 K and 1 atm pressure. At 273.1 K, the molar heat of fusion of ice, DeltaH_(f) is 6.00 kJ mol^(-1) and at 373.1 K, the molar heat of vapourization of water, DeltaH_(v) , is 40.6 kJ mol^(-1) . Also assume that the molar heat capacities, C_(p) , in the temperature range 373.1 to 273.1 K remains constant. Given that C_(p) = 75.25 m mol^(-1)K^(-1) and log 13.66 = 1.1354.