Diborane (B2H6) has been considered as a possible rocket fuel. Calculate DH° for the reaction B2H6(g) ---> 2B(s) + 3H2(g) using the following data: DH° (kJ) 2B(s) + 1.5O2(g) --> B2O3(s) –1273 B2H6(g) + 3O2(g) --> B2O3(s) + 3H2O(g) –2035 H2(g) + O2(g) --> H2O(g) –242 A. DH° = –36 kJ B. DH° = +36 kJ C. DH° = –86 kJ D. DH° = –86 kJ

A. DH° = –36 kJ

Explanation:

It is possible to obtain DH° of a reaction by the sum of DH° of half reactions. The DH° of the reaction:

B₂H₆(g) → 2B(s) + 3H₂(g)

Could be obtained from:

(1) 2B(s) + 1.5O₂(g) → B₂O₃(s) DH° = –1273kJ

(2) B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) DH° = –2035kJ

(3) H₂(g) + 0.5O₂(g) → H₂O(g) DH° = –242kJ

The sum of (2) - (1) gives:

B₂H₆(g) + 1.5O₂(g) → 2B(s) + 3H₂O(g) DH° = -2035kJ - (-1273kJ) = -762kJ

Now, this reaction - 3×(3):

B₂H₆(g) → 2B(s) + 3H₂(g) DH° = -762kJ - (3×-242kJ) = -36kJ

Thus, right answer is:

A. DH° = –36 kJ

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.