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KatRina [158]
3 years ago
11

Balance the following equations:

Chemistry
1 answer:
lesya [120]3 years ago
7 0
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

(I) N_{2} + 3H_{2} --> 2NH_{3}

(II) P_{4}  + 5O_{2}  --> 2P_{2} O_{5}

(III) 2NaF + Br_{2}  -->2NaBr + F_{2}

(IV) 2ZnS + 3O_{2} --> 2ZnO + 2SO_2

(V) Pb(NO_3)_2 + 2NaCl --> 2NaNO_3 + PbCl_2

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

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Consider the following equilibrium systems: A⇌2B ΔH o =20.0 kJ/mol Reaction 1 A+B⇌C ΔH o =−5.4 kJ/mol Reaction 2 A⇌B ΔH o =0.0 k
galina1969 [7]

Answer:

Reaction 1: Kc increases

Reaction 2: Kc decreases

Reaction 3: The is no change

Explanation:

Let us consider the following reactions:

Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol

Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol

Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol

To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>

Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.

Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.

Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.

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