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Free_Kalibri [48]
3 years ago
7

1) The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K.

Chemistry
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

The enthalpy of the reaction is 64.9 kJ/mol.

Explanation:

H_2 + \frac{1}{2}O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ..[1]

CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ..[2]

C_2H_6 + \frac{7}{2}O_2\rightarrow 2CO_2 + 3H_2O,\Delta H_3= -1559.9 kJ..[3]

2CH_4(g)\rightarrow C_2H_6(g) + H_2(g),\Delta H_4=?..[4]

2 × [2] - [1]- [3] = [4]  (Using Hess's law)

\Delta H_4=2\times \Delta H_2 -\Delta H_1 -\Delta H_3

\Delta H_4=2\times (-890.4 kJ)-(-285.8 kJ) -(-1559.9 kJ)

\Delta H_4=64.9 kJ/mol

The enthalpy of the reaction is 64.9 kJ/mol.

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Answer:

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Explanation:

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At same volume, for two situations, the above equation can be written as:-

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Given ,  

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The conversion of T( °C) to T(K) is shown below:

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So,  

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\frac{{n_1}\times {T_1}}{P_1}=\frac{{n_2}\times {T_2}}{P_2}

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