Question

The standard enthalpy of reaction for the given reaction is given as $xkJ$:
$B\left(s\right)+\frac{3}{4}{O}_2\left(g\right)\to \frac{1}{2}{B}_2{O}_3\left(s\right)$
Then the standard enthalpy of reaction for the reaction given below is equal to:
$4B\left(s\right)+3O_2\left(g\right)\to 2B_2{O}_3\left(s\right)$

• A. $xkJ$
• B. $2xkJ$
• C. $4xkJ$
• D. None of these

Answer 1

The correct option is C $4xkJ$

For the reaction,
$B\left(s\right)+\frac{3}{4}{O}_2\left(g\right)\to \frac{1}{2}{B}_2{O}_3\left(s\right)$
$△H_r^o=\frac{1}{2}△H_f^o(B_2{O}_3,s)-△H_f^o(B,s)-\frac{3}{4}△H_f^o(O_2,g)=xkJ$

multiplying the above reaction by 4 we get :
$4B\left(s\right)+3O_2\left(g\right)\to 2B_2{O}_3\left(s\right)$
so, the enthalpy change will also get 4 times ,
$△H_r^o=2△H_f^o(B_2{O}_3,s)-4△H_f^o(B,s)-3△H_f^o(O_2,g)=4xkJ$

Standard enthalpy of reaction:
Enthalpy change for a reaction when all the reactants and products are in their standard states.

This implies that reactants and products are taken at 1 bar pressure and usually at temperature of 298 K.