Question

An imaginary reaction $X\to Y$ takes place in three steps $X\to A,\Delta H=-q_1$; $B\to A,\Delta H=-q_2$; $Y\to B,\Delta H=-q_3$ If Hess' law is applicable, then the heat of the reaction $X\to Y$ is:

• A. $q_1-q_2+q_3$
• B. $q_2-q_1+q_3$
• C. $q_1-q_2-q_3$
• D. $q_3-q_2-q_1$

Answer 1

Answer: (B)

$q_2-q_1+q_3$

Solution:-

$X⟶A\Delta H=-q_1.....\left(1\right)$

$B⟶A\Delta H=-q_2$

$A⟶B\Delta H=q_2.....\left(2\right)$

$Y⟶B\Delta H=-q_3$

$B⟶Y\Delta H=q_3.....\left(3\right)$

Hess's Law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

Therefore,

Adding ${eq}^n\left(1\right),\left(2\right)\&\left(3\right)$, we have

$X+A+B⟶A+B+Y\Delta H=(-q_1)+\left(q_2\right)+\left(q_3\right)$

$X⟶Y\Delta H=q_2-q_1+q_3$

Hence the heat of the reaction $X\to Y$ is $q_2-q_1+q_3$.