Question

$\Delta H_{f\left(x\right)},\Delta H_{f\left(y\right)},\Delta H_{f\left(R\right)}$ and $\Delta H_{f\left(S\right)}$ denote the enthalpies of formation of $x,y,R$ and $S$ respectively. The enthalpy of the reaction $x+y\to R+S$ is:

• A. $\Delta H_{f\left(x\right)}+\Delta H_{f\left(y\right)}$
• B. $\Delta H_{f\left(R\right)}+\Delta H_{f\left(S\right)}$
• C. $\Delta H_{f\left(x\right)}+\Delta H_{f\left(y\right)}-\Delta H_{f\left(R\right)}-\Delta H_{f\left(S\right)}$
• D. $\Delta H_{f\left(R\right)}+\Delta H_{f\left(S\right)}-\Delta H_{f\left(x\right)}-\Delta H_{f\left(y\right)}$

Answer 1

The correct option is D $\Delta H_{f\left(R\right)}+\Delta H_{f\left(S\right)}-\Delta H_{f\left(x\right)}-\Delta H_{f\left(y\right)}$

For any reaction the enthalpy change of the reaction is given by,
$\Delta H=\text{sum of heat of formation of products in stoichiometric ratio - sum of heat of formation of reactants in stoichiometric ratio}$
i.e, for any reaction
$aA+bB\to cC+dD$
$\Delta H=c\Delta H_{f\left(C\right)}+d\Delta H_{f\left(D\right)}-a\Delta H_{f\left(A\right)}-b\Delta H_{f\left(B\right)}$
Hence, for the given reaction, $x+y\to R+S$,
$\Delta H=\Delta H_{f\left(R\right)}+\Delta H_{f\left(S\right)}-\Delta H_{f\left(x\right)}-\Delta H_{f\left(y\right)}$
Hence, answer is option $D$.