# 化學英文計算

1.

Urea, (NH2)2CO, can be prepared by heating ammonium cyanate, NH40CN.

NH4OCN變成(NH2)2CO

This reaction may occur by the following mechanism: NH4+ + OCN變成( NH) +HOCN fast eqllilibrium

NH3 + HOCN 變成( NH2)2CO slow

What is the rate law predicted by this mechanis

2. When calciwn carbonate, CaCO), is heated, it decomposes to calcium oxide. CaC03(s)→ CaO(s)+CO2(g);＊熱H=178.3kJ.

How much heat is required to decompose 10 g of calcium carbonate? (atornic weight: Ca = 40, C = 12, 0 = 16)

3.A water-soluble compound of gold and chlorine is treated with silver nitrate, AgN03, to convert the chlorine completely to silver chloride, AgCl. In an experiment, 328 mg of the compound gave 464 mg of silver chloride. Calculate the percentage of Cl in the compound. What is its empirical formula? (atomic weight: Cl = 35.5, N = 14, Ag = 108, Au = 197, O=16).

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• 8 years ago

1.

The reaction : NH4OCN → (NH2)2CO

Mechanism :

Step 1 : NH4⁺ + OCN⁻ ⇌ NH3 + HOCN (equilibrium constant = Ka)(fast)

Step 2 : NH3 + HOCN → (NH2)2CO (slow)

From step 1 :

Ka = [NH3] [HOCN] / [NH4⁺] [OCN⁻]

Hence, [NH3] [HOCN] = Ka [NH4⁺] [OCN⁻] ...... (1)

From step 2 :

Rate = k' [NH3] [HOCN] ...... (2)

Substitute (1) into (2) :

Rate = k' Ka [NH4⁺] [OCN⁻]

Hence, Rate = k [NH4⁺] [OCN⁻]

2.

CaCO3(s) → CaO(s) + CO2(g) .. ΔH = +178.3 kJ

When 1 mol of CaCO3 is decomposed, 178.3 kJ of heat is required.

No. of moles of CaCO3 = 10/(40 + 12 + 16x3) = 0.1 mol

Amount of heat required = 178.3 x 0.1 = 17.83kJ

3.

Mass fraction of Cl in AgCl = 35.5/(108 + 35.5) = 35.5/143.5

Mass of Cl in 464 g of AgCl = 464 x (35.5/143.5) = 114.8 mg

In 328 mg of the compound :

Mass of Cl = 114.8 mg

Mass of Au = 328 - 114.8 = 213.2 mg

In the compound, mole ratio Au : Cl

= 213.2/197 : 114.8/35.5

= 1.08 : 3.23

≈ 1 : 3

Hence, empirical formula = AuCl3

Source(s): 賣女孩的火柴