- E(Bonds broken) = 1371 kJ/mol reaction
- E(Bonds formed) = 1852 kJ/mol reaction
- ΔH = -481 kJ/mol.
- The reaction is exothermic.
<h3>Explanation</h3>
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:
According to the Henderson-Hasselbalch equation, the relation between pH and
is as follows.
pH = 
where, pH = 7.4 and
= 7.21
As here, we can use the
nearest to the desired pH.
So, 7.4 = 7.21 + 
0.19 = 
= 1.55
1 mM phosphate buffer means
+
= 1 mM
Therefore, the two equations will be as follows.
= 1.55 ............. (1)
+
= 1 mM ........... (2)
Now, putting the value of
from equation (1) into equation (2) as follows.
1.55
= 1 mM
2.55
= 1 mM
= 0.392 mM
Putting the value of
in equation (1) we get the following.
0.392 mM +
= 1 mM
= (1 - 0.392) mM
= 0.608 mM
Thus, we can conclude that concentration of the acid must be 0.608 mM.