asked in 科學化學 · 1 decade ago

化學計算五題,請大家幫幫忙

1.將NaCl(aq) 通過含某種離子交換樹脂的管柱後,溶液pH值將會增大。下列敘述何者正確?

(A)將KNO3(aq) 通過此離子交換樹脂的管柱後,其溶液pH值會減小

(B)將K2CrO4(aq) 通過含此離子交換樹脂的管柱後,其溶液明顯變色

(C)此離子交換樹脂可用來軟化硬水

(D)此離子交換樹脂可用濃HCl(aq) 沖洗,使其再生

2. Which of the following statements is (are) true?

a.Enthalpy is a state function.

b.In exothermic reactions, the reactants are lower in potential energy than the products.

c.A chemist takes the surroundings point of view when determining the sign for work or heat.

d.The heat of reaction and change in enthalpy can always be used interchangeably.

答案a,

我想請問c是什麼意思,為什麼不對?

3.For which process is ΔS negative?

a.evaporation of 1 mol of CCl4(l)

b.mixing 5 mL ethanol with 25 mL water

c.compressing 1 mol Ne at constant temperature from 1.5 atm to 0.5 atm

d.raising the temperature of 100 g Cu from 275 K to 295 K

答案是c,為什麼這裡的亂度是負的?

4. The reaction below occurs in basic solution. In the balanced equation, what is the sum of the coefficients?

Zn + NO3– → Zn(OH)42– + NH3

a.12

b.15

c.19

d.23

答案:d

我平衡完沒答案,請大家幫我看一下,哪裡錯了?

13H2O + 4Zn + 1NO3(-) → 4Zn(OH)4(2-) + NH3 + 7H(+)

5.The reaction

N2(g) + 3H2(g) 2NH3(g)

takes place in a 1.0-L vessel initially filled with an equimolar mixture of N2 and H2 at a pressure of 12.0 atm. When equilibrium is reached, the partial pressure of H2 is 3.0 atm. Determine the value of Kp for this reaction.

a.0.915

b.0.333

c.0.133

d.0.030

答d

Update:

感謝 豆漿 的解惑,這兩題我懂了^o^

2 Answers

Rating
  • 小白
    Lv 7
    1 decade ago
    Favorite Answer

    1. 通過離子交換樹脂後,水溶液的pH升高,表示陰離子被取代成OH-,亦即NaCl --> NaOH

    (A) KNO3 --> KOH ==> pH變大

    (B) K2CrO4 --> KOH ==> 黃色變無色,有明顯變色

    (C) 硬水是指水中含鈣離子或鎂離子,都是陽離子,此樹脂只能取代陰離子,不能軟化硬水

    (D) 此樹脂應用高濃度鹼性溶液再生

     

    2. 我們主要關心的是系統變化而不是外界變化。

     

    3. 若將氣體由V1壓縮成V2 (V2 < V1),則

    delta S = R x ln(V2/V1) < 0

     

    4. 題目中有"basic solution",不是基本溶液,而是鹼性溶液(我在耍冷)

    在鹼性溶液中不可能生成H+,你的平衡方程式移項一下就對了,H2O - H+ = OH-

    改寫為:4Zn + NO3 - + 6H2O + 7OH- --> 4Zn(OH4) 2- + NH3

    係數和 = 4+1+6+7+4+1 = 23

     

    5. 原本PH2 = PN2 = 6 atm,後來PH2變成3 atm,減少3 atm,由計量方程式,知PN2減少1atm,成為5 atm,而生成PNH3 = 2atm。

    Kp = PNH3^2 / ( PN2 x PH2^3 ) = 2^2 / ( 5 x 3^3 ) = 4/135 = 0.03

     

  • 1 decade ago

    (我先回答兩題 之後待續)

    2.根據IUPAC定義:w = -Pext dv (壓力必須是外壓)

    根據此定義,一系統膨脹時,對外界做功,IUPAC定義下壓力為負值

    功是一種能量的形式(單位為joule,雖然不是state function),

    這個負值的物理意義就代表了我們此時關心的是反應系統的能量變化(重要!)

    因為我們關心反應系統,一個系統對外界做功代表系統本身的能量轉移至環境

    因此系統能量下降,以功的形式把能量轉出去了。故對系統的能量變化是減小的。

    所以"A chemist takes the surroundings point of view when determining the sign for work or heat."的問題在於"surrounding"應改為"system",即"化學家由系統的觀點決定功或熱的正負號"而非由環境的觀點決定。

    3.compression(壓縮)基本上是體積變小,壓力變大的過程,熵變化應該小於零

    但是根據題意,1.5atm to 0.5atm,不是壓縮而是膨脹(expansion)

    所以熵變化根據壓縮小於零,根據數據是大於零,懷疑是題目出錯了。

    暫時先寫到這 晚點再補

    Source(s): 自己
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