Question

A reaction proceeds through two paths I and II to convert $X \to Z$. What is the correct relationship between, Q, ${Q_1}$, and ${Q_2}$ (Q represents a change in internal energy, here)?

A. $Q = {Q_1} \times {Q_2}$
B. $Q = {Q_1} + {Q_2}$
C. $Q = {Q_2} - {Q_1}$
D. $Q = \dfrac{{{Q_1}}}{{{Q_2}}}$

Answer 1

We know that the Internal energy is a state function. In the question above, Q is the change in internal energy. We all know that,
$\delta Q = \;\delta U + \delta W$.

Complete step by step answer:
We can define Internal Energy as the energy that is random and disordered motion of the molecule. To make it clear, we should know that the change in internal energy for a reaction is always the same whether it occurs in one step or the reaction either occurs in two steps or is in multiple steps that is in a series of steps. So in the above reaction, if X is converted to Z in single step, then the internal energy that is Q released and when the change happens through two step that is X to and then Y to Z, then the internal energy release while change from X to Y that is ${Q_1}$ and the internal energy released from change while Y to Z that is Q will definitely be equal to the energy while direct conversion of X to Z. so ${Q_1}$ that is released while conversion of X to Y and ${Q_2}$ that is released while conversion from Y to Z will be equal to Q that is released from conversion of X to Z.
Internal energy (Q) is a state function.
$\Rightarrow {Q_{XZ}} = {Q_{XY}} + {Q_{YZ}}$
$\Rightarrow Q = Q1 + Q2$

Therefore, the correct answer is option (B).

Note: The internal energies released in a multi-step reaction will be equal to when the reaction happens in just one step. The enthalpy of reaction is defined as the internal energy of the system.