Question

Рассчитайте кол-во теплоты, выделяющейся при сгорании 5,6л водорода (объем измерен при нормальных условиях) с образованием H2O(ж)

Answer 1

4H2+O2 = 2H2O

Q= (2*0+2*0)-2*(-241.8)=+483.6(при н.у.)

5,6/89,6=х/483,6

х=30,225 кДж

 

 

 

 

 

 

 

это решение по термохимичекой функции состояния системы

 

 

 

 

 

 

 

 

Answer 2

H2+1/2O2=H2O   ∆r1H=286 кдж

 

1) n(H2)=V(H2)/V=5.6/22.4=0.25

 

2) n(H2)=n(H2O)=0.25

 

3) Q=∆r1H*n(H2O)=286*0.25=71.5 кдж 

Answer 3

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To answer your question, I will use the following data and formula:

- The molar volume of an ideal gas at normal conditions is 22.4 L/mol. - The molar mass of hydrogen is 2.016 g/mol. - The enthalpy of combustion of hydrogen is -285.8 kJ/mol.

The amount of heat released by the combustion of 5.6 L of hydrogen is calculated as follows:

$$n = \frac{V}{V_m} = \frac{5.6}{22.4} = 0.25 \text{ mol}$$

$$m = n \times M = 0.25 \times 2.016 = 0.504 \text{ g}$$

$$Q = n \times \Delta H_c = 0.25 \times (-285.8) = -71.45 \text{ kJ}$$

Therefore, the combustion of 5.6 L of hydrogen produces 0.504 g of water and releases 71.45 kJ of heat.Ответь+подробно.+Рассчитайте+кол-во+теплоты,+выделяющейся+при+сгорании+5,6л+водорода+(объем+измерен+при+нормальных+условиях)+с+образованием+H2O(ж)