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Question: Methanol can be prepared synthetically by heating carbon monoxide and hydrogen gases under pressure ...

Methanol can be prepared synthetically by heating carbon monoxide and hydrogen gases under pressure in the presence of a catalyst. The reaction is CO2(g)+2H2(g)−>CH3OH(l)C{{O}_{2}}(g)+2{{H}_{2}}(g)->C{{H}_{3}}OH(l). The enthalpy of this reaction by an appropriate combination of the following data:
Cgraphite+12O2(g)−>CO(g)      △H=−110.5 kJ mol−1 Cgraphite+O2(g)−>CO2(g)        ΔH=−393.5 kJ mol−1 H2(g)+12O2(g)−>H2O(l)      ΔH=−285.9 kJ mol−1 CH3OH(l)+32O2(g)−>CO2(g)+2H2O(l)    △H=−726.6 kJ mol−1 \begin{aligned} & {{C}_{graphite}}+\dfrac{1}{2}{{O}_{2}}(g)->CO(g)\,\,\,\,\,\,\vartriangle H=-110.5\,kJ\,mo{{l}^{-1}} \\\ & {{C}_{graphite}}+{{O}_{2}}(g)->C{{O}_{2}}(g)\,\,\,\,\,\,\,\,\Delta H=-393.5\,kJ\,mo{{l}^{-1}} \\\ & {{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g)->{{H}_{2}}O(l)\,\,\,\,\,\,\Delta H=-285.9\,kJ\,mo{{l}^{-1}} \\\ & C{{H}_{3}}OH(l)+\dfrac{3}{2}{{O}_{2}}(g)->C{{O}_{2}}(g)+2{{H}_{2}}O(l)\,\,\,\,\vartriangle H=-726.6\,kJ\,mo{{l}^{-1}} \\\ \end{aligned}
A. −128.2 kJ mol−1-128.2\,kJ\,mo{{l}^{-1}}
B. −121.4 kJ mol−1-121.4\,kJ\,mo{{l}^{-1}}
C. +128.2 kJ mol−1+128.2\,kJ\,mo{{l}^{-1}}
D. +121.4 kJ mol−1+121.4\,kJ\,mo{{l}^{-1}}

Explanation

Solution

Try to make use of the four reactions given; manipulate them to get the desired reaction that is given in the question. Accordingly add or subtract the values to get the desired answer for the question.

Complete answer:
In order to solve the question, let us find out what enthalpy of a reaction actually is. Enthalpy can be referred to as the energy a reaction is giving/ taking. In a reaction, enthalpy is not only the amount of heat generated, but also the product of the pressure and volume. So, both of these terms are added in order to find out the enthalpy. It is a state function and is different for each and every reaction. Now there are some rules regarding enthalpy in reactions.
Whenever a reaction’s number of moles of both the reactant and product side are multiplied or divided by a constant, then the enthalpy also gets multiplied or divided by the same constant, respectively.
Chemical reactions can be treated as mathematical equations. So, it is possible to add and subtract the reactions whose enthalpies are known and find out the enthalpy of an unknown reaction. As reactions get added or subtracted, their enthalpies also get added and subtracted respectively.
In our question, we have been given four equations, and our target equation is: CO2(g)+2H2(g)−>CH3OH(l)C{{O}_{2}}(g)+2{{H}_{2}}(g)->C{{H}_{3}}OH(l)
In order to achieve this equation we can manipulate the four equation in the following way:
−equation(1) + 2×equation(3) + equation(2) − equation(4)-equation(1)\,+\,2\times equation(3)\,+\,equation(2)\,-\,equation(4)
As the enthalpies also get added and subtracted, so the final enthalpy of the reaction is :
△H=110.5−571.8−393.5+726.6= −128.2 kJ mol−1\vartriangle H=110.5-571.8-393.5+726.6=\,-128.2\,kJ\,mo{{l}^{-1}}

So, we get our correct answer as option A.

Note:
Enthalpy is a state function, and not a path function. So, it is independent of the path that is taken for reaction to happen. Also, enthalpy is never an absolute value, by enthalpy, we mean the enthalpy change or the difference between final and initial enthalpy. It is so because it is not possible to find the absolute zero or the zero point of enthalpy.