Question

Choose the reaction in which \[\Delta H\] is not equal to $\Delta U$.
A. $C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)$
B. ${{C}_{2}}{{H}_{4}}(g)+{{H}_{2}}(g)\to {{C}_{2}}{{H}_{6}}(g)$
C. $2C(graphite)+{{H}_{2}}(g)\to {{C}_{2}}{{H}_{2}}(g)$
D. ${{H}_{2}}(g)+{{I}_{2}}(g)\to 2HI(g)$
E. ${{N}_{2}}(g)+{{O}_{2}}(g)\to 2NO(g)$

Answer 1

Hint: In this question, mole is defined as the entity which is used to calculate the larger value of small ions, atoms and molecules. Change in enthalpy is defined as the amount of heat evolved or absorbed during a reaction. \[\Delta H\] is not equal to $\Delta U$ when $\Delta {{n}_{g}}$ is not equal to zero.

Formula used:$\Delta H=\Delta U+\Delta {{n}_{g}}RT$
where, $\Delta H$ is the change in enthalpy, $\Delta U$ is the change in internal energy, $\Delta {{n}_{g}}$ is the change in number of moles in gaseous form, $R$ is the universal gas constant and $T$ is the temperature in Kelvin.
$\Delta {{n}_{g}}={{n}_{P}}-{{n}_{R}}$
where, ${{n}_{P}}$ and ${{n}_{R}}$ is used to represent the number of moles of gaseous product and reactant.

Complete step by step answer:
According to the formula,
$\Delta H=\Delta U+\Delta {{n}_{g}}RT$
where, $\Delta H$ is the change in enthalpy, $\Delta U$ is the change in internal energy, $\Delta {{n}_{g}}$ is the change in number of moles in gaseous form, $R$ is the universal gas constant and $T$ is the temperature in Kelvin.
If $\Delta {{n}_{g}}$ is not equal to zero, then only \[\Delta H\] is not equal to $\Delta U$.
Therefore, we need to calculate the change in the number of moles in gaseous form.
Now, we will check the value of $\Delta {{n}_{g}}$ from the given options.
A. $C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)$
\[\Delta {{n}_{g}}={{n}_{P}}-{{n}_{R}}\]
Now, we substitute the values in the given formula we get,
\[\Rightarrow \Delta {{n}_{g}}=1-1\]
\[\Rightarrow \Delta {{n}_{g}}=0\]
B. ${{C}_{2}}{{H}_{4}}(g)+{{H}_{2}}(g)\to {{C}_{2}}{{H}_{6}}(g)$
\[\Delta {{n}_{g}}={{n}_{P}}-{{n}_{R}}\]
Now, we substitute the values in the given formula we get,
\[\Rightarrow \Delta {{n}_{g}}=1-1\]
\[\Rightarrow \Delta {{n}_{g}}=0\]
C. $2C(graphite)+{{H}_{2}}(g)\to {{C}_{2}}{{H}_{2}}(g)$
\[\Delta {{n}_{g}}={{n}_{P}}-{{n}_{R}}\]
Now, we substitute the values in the given formula we get,
\[\Rightarrow \Delta {{n}_{g}}=1-2\]
\[\Rightarrow \Delta {{n}_{g}}=-1\]
D. ${{H}_{2}}(g)+{{I}_{2}}(g)\to 2HI(g)$
\[\Delta {{n}_{g}}={{n}_{P}}-{{n}_{R}}\]
Now, we substitute the values in the given formula we get,
\[\Rightarrow \Delta {{n}_{g}}=2-2\]
\[\Rightarrow \Delta {{n}_{g}}=0\]
E. ${{N}_{2}}(g)+{{O}_{2}}(g)\to 2NO(g)$
\[\Delta {{n}_{g}}={{n}_{P}}-{{n}_{R}}\]
Now, we substitute the values in the given formula we get,
\[\Rightarrow \Delta {{n}_{g}}=2-2\]
\[\Rightarrow \Delta {{n}_{g}}=0\]
Therefore, in these given reactions $\Delta {{n}_{g}}$ is not equal to zero in option (B). Hence, change in enthalpy in noy equal to change in internal energy.

So, the correct answer is Option B.

Note: Change in enthalpy is defined as the amount of heat absorbed or evolved in a reaction. It is denoted with the symbol, $\Delta H$. This term is only applicable for the reaction at constant pressure.
This formula is used in the first law of thermodynamics.
To calculate the change in number of moles, products and reactants should be in gaseo