Toby 發問時間: 科學化學 · 1 0 年前

兩題化學熱函問題~

第一題

A quantity of 1*10^2mL of 0.5M HCl was mixed with 1*10^2mL of

0.5M NaOH in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and NaOH solution was the same,

22.5度,and the final temperature of the mixed solution was 25.86度.

Calculate the heat change for the neutralization reaction on a molar basis.

NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)

Assume that the densities and specific heats of the solutions are the same

as for water(1g/mL and 4.184J/g*度,respectively).

第二題

用上面的方法算

A quantity of 4*10^2mL of 0.6M HNO3 is mixed with 4*10^2mL of 3M

Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity.

The initial temperature of both solutions is the same 18.46度. What is the

final temperature of the solution?

希望順便幫忙翻譯一下

另外 neutralization reaction on a molar basis 這句是什麼意思!?

已更新項目:

沒人會熱焓嗎!

1 個解答

評分
  • 光弟
    Lv 7
    1 0 年前
    最佳解答

    1.

    100 ml 0.5M HCl 與100 mL 0.5M NaOH 在恆壓熱卡計中混合, 卡計吸收之熱量可忽略, HCl and NaOH溶液開始溫度均為22.5 oC, 混合後變為25.86 oC.

    計算每莫耳酸鹼中和反應之熱變化.

    設溶液之密度與比熱, 與水相同.

    100 ml 0.5M HCl 含酸莫耳數 = 100 mL 0.5M NaOH含鹼莫耳數

    = 0.5 M x 0.1 L = 0.05 mol

    反應熱 = 200 ml x 1 g/ml x 4.184J/g-度 x (25.86度 - 22.5度) = 2815 J

    莫耳反應熱 = 2815 J / 0.05 mol = 56300 J/mol = 56.3 kJ/mol

    2.

    400 ml 0.6M HNO3與400 ml 3M Ba(OH)2, 在恆壓熱卡計中混合, 卡計吸收之熱量可忽略, 溶液開始溫度均為22.5 oC, 混合後變為?

    400 ml 0.6M HCl 含酸莫耳數 = 0.6 M x 0.4 L = 0.24 mol

    400 mL 3M Ba(OH)2含鹼莫耳數 = 2.4 mol

    產生熱 = 56300 J/mol x 0.24 mol = 13152 J

    增加溫度 = 13152 J / 4.184J/g-度 / (800 ml x 1 g/ml) = 3.93 度

    混合後變為: 22.5 + 3.93 = 26.43度

    the heat change for the neutralization reaction on a molar basis.- - - -

    ㄧ莫耳酸和ㄧ莫耳鹼中和反應熱.

    參考資料: 普化
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