the synthesis reaction to form boron hydride is: 2b(s) + 3h₂(g) → b₂h₂(g), δh = +36 kj. what is the enthalpy…

the synthesis reaction to form boron hydride is: 2b(s) + 3h₂(g) → b₂h₂(g), δh = +36 kj. what is the enthalpy value for the manipulated reaction: 1/2b₂h₂(g) → b(s) + 3/2h₂(g)? -18 kj -36 kj +18 kj +72 kj done

the synthesis reaction to form boron hydride is: 2b(s) + 3h₂(g) → b₂h₂(g), δh = +36 kj. what is the enthalpy value for the manipulated reaction: 1/2b₂h₂(g) → b(s) + 3/2h₂(g)? -18 kj -36 kj +18 kj +72 kj done

Answer

Explanation:

Step1: Reverse the original reaction

When a reaction is reversed, the sign of $\Delta H$ changes. The original reaction is $2B(s)+3H_2(g)\rightarrow B_2H_2(g), \Delta H = + 36\ KJ$. The reversed reaction is $B_2H_2(g)\rightarrow2B(s)+3H_2(g),\Delta H=-36\ KJ$.

Step2: Scale the reversed - reaction

The manipulated reaction is $\frac{1}{2}B_2H_2(g)\rightarrow B(s)+\frac{3}{2}H_2(g)$. We are scaling the reversed reaction by a factor of $\frac{1}{2}$. When a reaction is scaled by a factor $n$, the $\Delta H$ is also scaled by the same factor. So $\Delta H=\frac{-36}{2}=-18\ KJ$.

Answer:

-18 KJ