Question

Pentaborane b5h9(s) burns vigorously in o2 to give b2o3(s) and h2o(l). what is δh° for the combustion of 1 mol of b5h9(s)? substance δh°f (kj/mol) b2o3(s) –1273.5 b5h9(s) +73.2 h2o(l) –285.8

Answer 1

Write  a   balanced  chemical  equation
2B5H9  +  12O2  ---> 5B205  +9 H20
From  the  Hess  law  wich  state  that   regardless  of  the multiple   steps  of  chemical  reaction  the  total  enthapy  is  equal  to  the  sum  of  all  changes
hence;
{(5  x  -1273.5)+(9  x  -289.8)}  - { (73.2  x2)  +(0  x12)}=  -9086.1kj  /mol

Answer 2

Answer: The enthalpy of burning of 1 mole of $B_5H_9(s)$ is -4543.05 kJ/mol

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

For the given chemical reaction:

$2B_5H_9(s)+12O_2\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3)})+(9\times \Delta H^o_f_{(H_2O)})]-[(2\times \Delta H^o_f_{(B_5H_9)})+(12\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(B_2O_3)}=-1273.5kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_f_{(H_2O)}=-285.8kJ/mol\\\Delta H^o_f_{(B_5H_9)}=73.2kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times (-1273.5))+(9\times (-285.8))]-[(2\times (73.2))+(12\times 0)]\\\\\Delta H^o_{rxn}=-9086.1kJ/mol$

The enthalpy of burning of 2 moles of $B_5H_9(s)$ is coming out to be -9086.1 kJ/mol

So, enthalpy of burning of 1 mole of $B_5H_9$ will be = $\frac{-9086.1kJ/mol}{2}\times 1=-4543.05kJ/mol$

Hence, the enthalpy of burning of 1 mole of $B_5H_9(s)$ is -4543.05 kJ/mol