Question

$(\Delta H-\Delta U)$ for the formation of carbon monoxide (CO) from its element at 298K is: $(R=8.314J{{K}^{-}}mo{{l}^{-}})$
(A) -1238.78$Jmo{{l}^{-}}$
(B) 1238.78$Jmo{{l}^{-}}$
(C) -2477.57$Jmo{{l}^{-}}$
(D) 2477.57$Jmo{{l}^{-}}$

Answer 1

Hint: The relation between change in enthalpy and internal energy with the temperature of the reaction system is given below.
$\Delta H=\Delta U+\Delta nRT$
One mole of carbon element reacts with one half a mole of oxygen gas to produce one mole of carbon monoxide gas.

Complete step by step answer:
Let’s first write the reaction of production of carbon monoxide from its elements.
\[{{C}_{(s)}}+\dfrac{1}{2}{{O}_{2(g)}}\to C{{O}_{(g)}}\]
Now, we are given to find the value of $\Delta H-\Delta U$ of the reaction. So, we need to put a formula which relates enthalpy, internal energy and temperature of the reaction. This formula is given as under:
We can write that H = U + PV
But for any gas, we can write that PV = nRT where n is number of moles of gas. So, as we put this value in above given equation, we get,
H = U + nRT…...(1)
Now, we need to find the changes in given quantities, so, we can rewrite the equation (1) as
     $\Delta H=\Delta U+\Delta nRT$ ……….(2)
So, we can write equation (2) as $\Delta H-\Delta U=\Delta nRT$ ………..(3)
Now, we are given that universal gas constant R = 8.314$J{{K}^{-}}mo{{l}^{-}}$
Temperature T = 298 K
And $\Delta n$ = Moles of gases in product – Moles of gases in reactants
So, from the reaction, we can write that $\Delta n=1-\dfrac{1}{2}=0.5moles$
Now, put all these values into equation (3).
     $\Delta H-\Delta U=0.5\times 8.314\times 298$
     $\Delta H-\Delta U=1238.786Jmo{{l}^{-}}$

Thus, we can conclude that option (B) is a correct answer for this question.

Note:
Make sure that you put the value of temperature in Kelvin unit these types of equations of thermodynamics. We will need to put the value of change in moles of gases during the reaction in the equation (2), so do not consider putting a total of all moles of gases.