Question

the standard enthalpy of combustion of solid boron numerically equal to​

Answer 1

Answer:

standard heat of combustion of solid boron is equal to one half the standard heat of formation of B2O3.

Answer 2

Correct question:

The standard enthalpy of combustion of solid boron is numerically equal to

 $a)\ \Delta_f H^0 (B_2O_3)\\ \\b)\ \frac{1}{2} \Delta_f H^0 (B_2O_3) \\ \\c)\ 2 \Delta_f H^0 (B_2O_3)\\ \\d)\ -\frac{1}{2} \Delta_f H^0 (B_2O_3)$

Solution:

To find:

The standard enthalpy of  combustion of solid boron

Calculation:

B₂O₃ is obtained by the reaction of solid boron(B) and oxygen. The balanced chemical reaction is given is

     $2B+\frac{3}{2} O_2\longrightarrow B_2O_3$

As we need the enthalpyof combustion for one mole of B, the reaction for one mole of combustion of B is

     $B+\frac{3}{4} O_2\longrightarrow \frac{1}{2} B_2O_3$

From the reaction it can be concluded that

      $\Delta_c H^0\ of\ boron =\frac{1}{2} \Delta_f H^0\ of\ B_2O_3$

The standard enthalpy of  combustion of solid boron is numerically equal to   $\frac{1}{2} \Delta_f H^0(B_2O_3)$