Answer:
The standard enthalpy change for the reaction at is -2043.999kJ
Explanation:
Standard enthalpy change () for the given reaction is expressed as:
Where refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
Enthalpy is energy of bonds broken - energy of bonds formed. Here, the NH3 and O2 are broken and H2O and NO are formed. So the energy to break the NH3 bonds is 3 times the amount of energy it takes to break a N-H single bond (because there are three of them in a NH3 molecule) and then multiplied by 4 because there are four particles.
So the energy of the bonds broken is 12x the energy to break a N-H single bond plus 5x the amount of energy to break an O—O double bond (you don’t multiply this by anything because in each O2 molecule there is only one bond).
The energy of the bonds formed is 6*2 = 12 Times the amount of energy for a O-H single bond plus 4 times the amount of energy required to break a N—O double bond.
Subtract energy of bonds broken - energy of bonds formed and this is the change in enthalpy.
To know what type of bond it is, draw the Lewis structure.
Answer:
1. Dynamic equilibrium is a equilibrium in which the rate of forward direction is equal to the rate of backward direction. It is represented by a right left arrow.
2) Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
For the given chemical reaction:
The expression for is written as:
3) The value of is 1000 , which means the products are more favoured as compared to reactants and that the equilibrium lies more towards product side.
Answer: The solubility of this compound in pure water is 0.012 M
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as
The equation for the ionization of the is given as:
By stoichiometry of the reaction:
1 mole of gives 1 mole of and 2 mole of
When the solubility of is S moles/liter, then the solubility of will be S moles\liter and solubility of will be 2S moles/liter.
Thus solubility of this compound in pure water is 0.012 M