URGENT PLEASE HELP
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Answer : The correct option is, (D) 3600 kJ
Explanation :
Mass of octane = 75 g
Molar mass of octane = 114.23 g/mole
Enthalpy of combustion = -5500 kJ/mol
First we have to calculate the moles of octane.
Now we have to calculate the heat released in the reaction.
As, 1 mole of octane released heat = -5500 kJ
So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)
= -3608 kJ
≈ -3600 kJ
Therefore, the heat released in the reaction is 3600 kJ
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:
A. C is added to the reaction mixture.
If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.
B. is condensed and removed from the reaction mixture.
If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.
C. CO is added to the reaction mixture.
If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.
D. is removed from the reaction mixture.
If the concentration of product specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of product specie occurs. So, on removing hydrogen , the equilibrium will shift in right direction.