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mart [117]
3 years ago
7

Why is ∆Hf ° 62.4 kJ/mol for I2 ( g., but 0.0 kJ/mol for I2 (s)?

Biology
2 answers:
Oksana_A [137]3 years ago
8 0

<u>Answer:</u> Because enthalpy of the substances in their standard state is always 0.

<u>Explanation:</u>

Standard enthalpy of formation or standard heat of formation is defined as the change in enthalpy during the formation of 1 mole of a substance from its constituent elements when all the substances are present in their standard states.

Iodine is the 53rd element which is present in Group 17 and Period 5 of the period table.

Standard enthalpy formation of the substances present in their standard state is always equal to 0.

As, iodine occurs in solid state only. So, its standard formation enthalpy will always be 0 in solid state.

Diano4ka-milaya [45]3 years ago
6 0
Opposite the ordinal<span> equation, the reaction </span>can<span> become </span>energy-absorbing<span>,
 N2 (g) + </span>four liquid<span> (l) ---> </span>a pair of<span> NO2 (g) + </span>four<span> H2 (g),

delta H = 212.5 kJ/mol </span>a pair of<span> NH3 (g) ----------------> N2 (g) + </span>three<span> H2 (g)
 delta H = </span>100<span> and fifteen kJ/mol </span>transfer<span> those </span>a pair of<span> equations at </span>constant<span> time.

You get the equation </span>needed and also the succeeding<span> delta H </span>is that the total<span> of the </span>several<span> 
delta H: 4H2O (l) + 2NH3 (g) ------> 2NO2 (g) + 7H2 (g) , delta H = 327.5 kJ/mo</span>
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