2 answers:
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- E(Bonds broken) = 1371 kJ/mol reaction
- E(Bonds formed) = 1852 kJ/mol reaction
- ΔH = -481 kJ/mol.
- The reaction is exothermic.
2 H-H + O=O → 2 H-O-H
There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb
- E(Bonds broken) = 2 × 436 + 499 = 1371 kJ/mol reaction.
- ΔH(Breaking bonds) = +1371 kJ/mol
Each mole of the reaction will form two moles of water molecules. Each mole of H₂O molecules have two moles O-H bonds. Two moles of the molecule will have four moles of O-H bonds. Forming all those bond will release
- E(Bonds formed) = 2 × 2 × 463 = 1852 kJ/mol reaction.
- ΔH(Forming bonds) = - 1852 kJ/mol
Heat of the reaction:
is negative. As a result, the reaction is exothermic.
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
For the given equation:
Given that reaction is an endothermic reaction.
For the given options:
a)Adding more C
If the concentration of C that is the reactant is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of C takes place. Therefore, the equilibrium will shift in the right direction to wards the formation of hydrogen gas.
b) Adding more
If the concentration of water that is the reactant is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of water takes place. Therefore, the equilibrium will shift in the right direction towards the formation of hydrogen gas.
c) Raising the temperature of the reaction mixture
If the temperature is increased,heat of the equilibrium mixture will also increase so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in heat that is decrease in temperature occurs.
As, this is an endothermic reaction, forward reaction will decrease the temperature. Hence, the equilibrium will shift in the right direction that is towards the formation of hydrogen gas.
d) Increasing the volume of reaction mixture
If the volume of the container is increased, the pressure will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where increase in pressure is taking place. As the number of moles of gas molecules is greater at the product side. So, the equilibrium will shift in the right direction that is towards the formation of hydrogen gas.
e) Adding a catalyst to reaction mixture
Role of catalyst is to attain the equilibrium quickly without disturbing the state of equilibrium. Hence, addition of catalyst will not change the equilibrium of the reaction.
f) Adding an inert gas to reaction mixture
Adding inert gas to the mixture at constant volume will not effect the equilibrium. Hence, addition of an inert gas will not change the equilibrium of the reaction.