請問幾題熱力學的題目

1. The standard enthalpy of formation of H2O(l) at 298K is -285kJ/mol. Calculate the change in internal energy for the following process at 298K and 1 atm:H2O(l)→H2(g) 1/2O2(g). △E°=?2. When water is supercooled

it freezes at a temperature below 0℃. If 10.9 kJ of heat is released when 2 mol of supercooled water at -15℃freezes

calculate the molar enthalpy of fusion for ice at 0℃ and 1 atm. Assume the molar heat capacities for H2O(s) and H2O(l) are 37.5J/Kmol and 75.3 J/Kmol

respectively

and are temperature independent.3. The heat of vaporization of water at the normal boiling point

373.2K

is 40.66kJ/mol. The specific heat capacity of liquid water is 4.184J/K‧g and for gaseous water is 2.2 J/K‧g. Assume that these values are independent of temperature. What is the heat of vaporization of water at 340.2K?
1. 由題目可知

液態水的標準生成熱(熵)為-285kJ/mol(298K)

即H2(g) 1/2O2(g)→H2O(l) △E°(△H°)= -285kJ/mol因此倒轉上列液態水生成反應式即與題目所問之反應式相同

此時的反應熱即為液態水生成熱乘以負號

為285kJ/mol。

2. -15℃液態水凝固的熱量變化(△H)為下列三步驟熱量變化的總和a. -15℃液態水昇溫至0℃(△Hh

吸熱

為正值)b. 0℃液態水凝固(△Hs

放熱

為負值)c. 0℃固態水(冰)降溫至-15℃(△Hc

放熱

為負值)題目所問之0℃固態水的熔解熱即為-△Hs(與凝固熱相差一負號)△H = △Hh △Hs △Hc-△Hs = △Hh △Hc - △H = 75.3*[0-(-15)] 37.5*(-15-0) – (-10.9*1000/2) = 6017 J/mol = 6.02 kJ/mol3. 340.2K液態水汽化的熱量變化(△H)為下列三步驟熱量變化的總和a. 340.2K液態水昇溫至373.2K (△Hh

吸熱

為正值)b. 373.2K液態水汽化(△Hv

吸熱

為正值)c. 373.2K氣態水(蒸汽)降溫至340.2K (△Hc

放熱

為負值)△H = △Hh △Hv △Hc= 18*4.184*(373.2-340.2) 40.66*1000 18*2.2*(340.2-373.2) = 41838.496 J/mol = 41.84 kJ/mol

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