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Question: Consider the following processes: | \(\text{ }\Delta \text{H (kJ /mol )}\) ---|--- \(\te...

Consider the following processes:

|  ΔH (kJ /mol )\text{ }\Delta \text{H (kJ /mol )}
---|---
 12A→B \text{ }\dfrac{1}{2}\text{A}\to \text{B }. |  +150 \text{ +150 }
 3B→ 2C + D \text{ 3B}\to \text{ 2C + D } |  −125 \text{ }-125\text{ }
 E + A → 2D \text{ E + A }\to \text{ 2D } |  +350 \text{ +350 }

For  B + D + E → 2C \text{ B + D + E }\to \text{ 2C } ,  (ΔH ) \text{ (}\Delta \text{H ) } will be:
A)  525 kJ/mol\text{ 525 kJ/mol}
B)  −175 kJ/mol\text{ }-175\text{ kJ/mol}
C)  −325 kJ/mol\text{ }-325\text{ kJ/mol}
D)  325 kJ/mol\text{ 325 kJ/mol}

Explanation

Solution

The enthalpy is the heat evolved or absorbed by the process. Here the enthalpy of a process can be found by arranging the process (by adding, subtraction, or multiplying by factors) to get the desired reaction. The enthalpies of the individual process are then treated the same which ultimately results in the enthalpy of the process.

Complete step by step answer:
The enthalpy is the heat evolved or absorbed by the process. Here the enthalpy of a process can be found by arranging the process (by adding, subtraction, or multiplying by factors) to get the desired reaction. The enthalpies of the individual process are then treated the same which ultimately results in the enthalpy of the process.

|  ΔH (kJ /mol )\text{ }\Delta \text{H (kJ /mol )}
---|---
 12A→B \text{ }\dfrac{1}{2}\text{A}\to \text{B }. |  +150 \text{ +150 }
 3B→ 2C + D \text{ 3B}\to \text{ 2C + D } |  −125 \text{ }-125\text{ }
 E + A → 2D \text{ E + A }\to \text{ 2D } |  +350 \text{ +350 }

We are interested to find out the enthalpy  (ΔH ) \text{ (}\Delta \text{H ) } for the reaction B + D → E + 2C \text{ B + D }\to \text{ E + 2C }.
The steps are as follows,

  1. Multiply the equation (1) by factor 2. We have,
     2×(12A→B ) = A → 2 B \text{ 2}\times \left( \dfrac{1}{2}\text{A}\to \text{B } \right)\text{ = A }\to \text{ 2 B } (i)
  2. Multiply the equation (3) by the factor of  −1 \text{ }-1\text{ } . We have,
     −1×(E + A → 2D ) ⇒ 2D → E + A \text{ }-1\times (\text{E + A }\to \text{ 2D ) }\Rightarrow \text{ 2D }\to \text{ E + A } (ii)
  3. Now add equation (i), (ii) and (3).We will get the desired equation.
     →2 \+3→2C +  \+2→E+    B + D→ E+ 2C ‾ \begin{aligned} & \text{ }\begin{matrix} {} & {\text{}} & \to & \text{2} \\\ \+ & \text{3} & \to & \text{2C + } \\\ \+ & \text{2} & \to & \text{E+} \\\ \end{matrix}\text{ } \\\ & \overline{\text{ B + D}\to \text{ E+ 2C }} \\\ \end{aligned}

Here, the A, B, and D cancel out from the product and reactant side.
4) Now the enthalpy of the process can be calculated as the same as the above. Here, we have to multiply the enthalpy of the equation (2) by the negative sign. Let’s calculate the enthalpy of the process,
 2×(150) \+(−125) \-(+350) −175 kJ/mol‾  \text{ }\begin{matrix} {} & 2\times (150) \\\ \+ & (-125) \\\ \- & (+350) \\\ {} & \overline{-175\text{ kJ/mol}} \\\ \end{matrix}\text{ }

The equation (1) is multiplied by factor 2, thus the enthalpy is also multiplied by factor 2. The enthalpy is equal to the addition of enthalpies of equation (1) and (2) and subtracts the (3) enthalpy from it.
Thus, the enthalpy of the process  B + D + E → 2C \text{ B + D + E }\to \text{ 2C } is equal to the −175 kJ/mol\text{ }-175\text{ kJ/mol}
So, the correct answer is “Option B”.

Note: When heat is absorbed by the solution that is cooler,  (ΔH ) \text{ (}\Delta \text{H ) } is given a positive sign. If the heat is evolved and given to the solution, that is the solution is warmer,  (ΔH ) \text{ (}\Delta \text{H ) } is given a negative sign.