Question

The synthesis reaction to form boron hydride is: 2B(s) + 3H, (g) – B,H,(9) AH = +36 KJ. what is the enthalpy value for the manipulated reaction: 1/ (g) b(s) 3/ (g)? a. –18 kj b. –36 kj c. 18 kj

Answer 1

The enthalpy value for the manipulated reaction is -36 kJ.

Step-by-step explanation:

The synthesis reaction to form boron hydride is: 2B(s) + 3H2 (g) -> B2H6 (g) + 36 kJ. In the manipulated reaction, the coefficients in the reaction equation have been changed. To determine the enthalpy value for the manipulated reaction, we need to multiply the original enthalpy value by the stoichiometric coefficient ratio, which is 1:3. Thus, the enthalpy value for the manipulated reaction is -36 kJ (option b).