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Sergio [31]
3 years ago
13

When are the products of an equalibrium reaction favored

Chemistry
1 answer:
Leviafan [203]3 years ago
4 0
For exothermic reaction (that is heat is considered as a product) then
 reactants= product + Energy ,

An decrease in temperature  favors the forward reaction (more product formed)

pressure depends on number of mole of the product and reactants

a A + b B + c C + d D (Where a b c and d are numerical value and A B C and D are compounds) :
if more A is added to the equilibrium mixture, equilibrium position will shift to the right (more product formed) so as to decrease concentration of A
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Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,ΔH° = ?

We are given with:

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ΔH° =  

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})

\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)

\Delta H^o=-794kJ

\Delta H^o>0 endothermic reaction

\Delta H^o exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

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