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What is the value of $\Delta H$ for the reaction $X+2Y\to 2Z$ ?
$W+X\to 2Y$, $\Delta H=-400Kcal/mol$
$2W+3X\to 2Z+2Y$,$\Delta H=\,-150Kcal/mol$
$\left( A \right)\,\,-550Kcal/mol$
$\left( B \right)\,\,+50\ Kcal/mol$
$\left( C \right)\,\,-50Kcal/mol$
$\left( D \right)\,\,+650Kcal/mol$
$\left( E \right)\,\,+250\ Kcal/mol$

Answer
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Hint: Enthalpy is a property of a thermodynamic system, that is a convenient state function preferred in many measurements in chemical, biological, and physical systems at a constant pressure. It is defined as the sum of the system’s internal energy and the product of its pressure and volume.

Complete step by step answer:
Enthalpy is a state function whose value for a particular system cannot be measured. We can, however measure enthalpy changes $(\Delta H)$
Since, $\Delta H={{H}_{final}}-\ {{H}_{initial}}$,
 the enthalpy change for a chemical reaction is found by:
$\Delta {{H}_{rxn}}={{H}_{products}}-{{H}_{reac\tan ts}}$
Where, $\Delta {{H}_{rxn}}=\text{Enthalpy of reaction or the heat of reaction}\text{.}$
Calculation of enthalpy change in given reaction:
$\Rightarrow X+2Y\to 2Z$
Multiplying the equation given in question by $2$ as
$\Rightarrow $$2\times \ \left( W+X\to 2Y \right)$, $\Delta H=\ -2\times 400Kcal/mol$
$\Rightarrow $$2W+2X\to 4Y\ ....(1)$, $\Delta H=\ -800Kcal/mol$
$\Rightarrow $\[2W+3X\to 2Z+2Y\,\,....(2)\], \[\Delta H=\ -150Kcal/mol\]
\[\text{Equation}(1)-\text{Equation}(2)\]
$2W+3X\to 2Z+2Y$, $\Delta H=\ -150$
$2W+2X\to 4Y,$ $\Delta H=\ -800$
$\ X\to \ -2Y+2Z$, $\Delta H=-150-\left( -800 \right)=\,-150+800=650Kcal/mol$
$\ \ X+2Y\ \to \ 2Z$, $\Delta H=\ +650\,Kcal/mol$

So, the correct answer is Option D.

Additional Information:
In an exothermic reaction: Energy flows out of the system. The flow of heat is negative $\left( - \right)$ for the exothermic system.
In an endothermic reaction: Energy flows into the system. The flow of heat is positive $\left( + \right)$ for the endothermic system.
The Enthalpy $\Delta H$ is the amount of heat released or absorbed when a chemical reaction occurs at a constant pressure.
Hess’s law is useful to find the overall enthalpy change in a reaction, it does not take into account the route or the number of steps followed in a reaction, it states that the overall enthalpy change will be the same if the reaction is completed in one step or in multiple steps.

Note: The enthalpy of a uniform system is defined as $h=\dfrac{H}{M}$ , where $M$ is the mass of the system. The $SI$ unit for specific enthalpy is Joule per kilogram or $J/Kg$.
It can be expressed in other specific quantities by $h=u+pV$ where $u$ is the specific internal energy, $p$ is the pressure and $V$ is the specific Volume, which is equal to $\dfrac{1}{p}$ , where $p$ is the density.