Considering the following reactions, what type of reaction is represented by these two examples?

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Considering the following reactions, what type of reaction is represented by these two examples?

$M g (s) + \frac{1}{2} O_{2} (g) \to M G O (s) + 146$ $Mg(s) + 1/2 O_2(g) -> MGO(s) + 146$ kcal/mol
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ $H_2(g) + 1/2 O_2(g) -> H_2O(g)$ $DeltaH = -57.82$ kcal/mol.

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Answer 1 · 2016-07-07T16:03:21

These are both $"enthalpies of formation"$ , $DeltaH_f^@$

Explanation:

The $"enthalpy of formation"$ is the enthalpy associated with the formation of one mole of product from its constituent elements under standard conditions.

Note that you have listed enthalpy as a product in the formation of $MgO$ ; of course you are aware that this represents a NEGATIVE $DeltaH_f^@$ . If you ar not clear on why I should mention this, ask for clarification.

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Considering the following reactions, what type of reaction is represented by these two examples?

$M g (s) + \frac{1}{2} O_{2} (g) \to M G O (s) + 146$ $Mg(s) + 1/2 O_2(g) -> MGO(s) + 146$ kcal/mol
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ $H_2(g) + 1/2 O_2(g) -> H_2O(g)$ $DeltaH = -57.82$ kcal/mol.

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Explanation:

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Considering the following reactions, what type of reaction is represented by these two examples?

Mg(s) + 1/2 O_2(g) -> MGO(s) + 146Mg(s)+12O2(g)→MGO(s)+146Mg(s) + 1/2 O_2(g) -> MGO(s) + 146 kcal/mol H_2(g) + 1/2 O_2(g) -> H_2O(g)H2(g)+12O2(g)→H2O(g)H_2(g) + 1/2 O_2(g) -> H_2O(g) DeltaH = -57.82DeltaH = -57.82 kcal/mol.

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$M g (s) + \frac{1}{2} O_{2} (g) \to M G O (s) + 146$ $Mg(s) + 1/2 O_2(g) -> MGO(s) + 146$ kcal/mol
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ $H_2(g) + 1/2 O_2(g) -> H_2O(g)$ $DeltaH = -57.82$ kcal/mol.