Given the following chemical reaction what is the bond enthalpy of OH 4NH3 3O2 yields 2N2 6H20 Change H 1266 kJ NH equals 389 O O equals 498 and N N equals 941 3
Volume of RBCs in venous circulation increased by 3% than than of RBCs in arterial circulation, why?
To determine the bond enthalpy of a specific bond in a chemical reaction, you need to know the bond enthalpies of the bonds in all the reactants and products involved.
In the given reaction:
4NH3 + 3O2 -> 2N2 + 6H2O
Change H = 1266 kJ
We are interested in the bond enthalpy of OH (hydroxide ion). However, it is not directly present in the given reaction. To find the bond enthalpy of OH, we need to break down the reaction into individual bond breaking and bond forming steps.
First, let's identify the bonds that are broken and formed in this reaction:
Bonds broken: N-H, N-H, N-H, N-H, O=O
Bonds formed: N≡N, N≡N, O-H, O-H, O-H, O-H, O-H, O-H
Now, we can calculate the overall bond enthalpy change in the reaction by summing up the bond enthalpies of bonds broken minus the bond enthalpies of bonds formed.
Change in bond enthalpy = (4 * N-H bond enthalpy) + (3 * O=O bond enthalpy) - (2 * N≡N bond enthalpy) - (6 * O-H bond enthalpy)
Given the bond enthalpies:
N-H = 389 kJ/mol (assuming this is for NH3)
O=O = 498 kJ/mol (for O2)
N≡N = 941 kJ/mol
O-H = unknown
Unfortunately, you didn't provide the bond enthalpy for O-H. Without this information, it's not possible to calculate the bond enthalpy of OH in this specific reaction.
To calculate the bond enthalpy of OH, you will need to determine the O-H bond enthalpy from a reliable source or data table. Once you have that value, you can substitute it into the equation mentioned earlier to find the bond enthalpy of OH.
Keep in mind that bond enthalpies can vary depending on the context and the specific molecules involved, so it is essential to use the appropriate bond enthalpy values for accurate calculations.