
For a reaction ${{A}_{(g)}}\to {{A}_{(l)}},\Delta H=-3RT$.The correct statement for the reaction is
A) $\Delta H=\Delta U=0$
B) $\left| \Delta H \right|\rangle \left| \Delta U \right|$
C) $\left| \Delta H \right|\langle \left| \Delta U \right|$
D) $\Delta H=\left| \Delta U \right|=0$
Answer
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Hint: Consider an equation that relates internal energy and enthalpy. The number of moles of gas present in the product side is zero and in the reactant side is 1.
Complete step by step solution:
In the question a reaction is given in which a gas, A is getting converted to liquid state.
The change in enthalpy for this reaction is, $\Delta H=-3RT$.Here the enthalpy value is negative and its means that it the system releases the energy to the surroundings. It is an exothermic process.
Now let’s write an equation which relates enthalpy ($\Delta H$) and the internal energy ($\Delta U$).
And the equation is, \[\Delta H=\Delta U+\Delta {{n}_{g}}RT\]
Here,\[\Delta H=\Delta U+\Delta {{n}_{g}}RT\]
\[\Delta {{n}_{g}}\]= number of moles of gas, T is temperature and R is gas constant
We calculate \[\Delta {{n}_{g}}\]=no. of moles of gas in product side – no. of moles of gas in reactant side
\[\Delta {{n}_{g}}\]=0-1=-1, here in the product side there are no gas molecules whereas there is 1 mole of gas in the reactant side.
Substitute the value of, \[\Delta {{n}_{g}}\] and $\Delta H$ in the above equation relating $\Delta H$ and $\Delta U$
Therefore the equation becomes,
$-3RT=\Delta U+\left( -1 \right)RT=-3RT=\Delta U-2RT$
$-2RT=\Delta U$
-2RT is greater than -3RT, hence\[\Delta U\rangle \Delta H\], change in internal energy is greater than change in enthalpy,
So from the above options the correct answer is option C, which is$\left| \Delta H \right|\langle \left| \Delta U \right|$
Note: $-2RT=\Delta U$
We can use the equation mentioned below to find internal energy parameters, change in number of moles of gas in a reaction, enthalpy.
\[\Delta H=\Delta U+\Delta {{n}_{g}}RT\]
Attention should be given for substituting the value of Since R can have different values in different units.
Complete step by step solution:
In the question a reaction is given in which a gas, A is getting converted to liquid state.
The change in enthalpy for this reaction is, $\Delta H=-3RT$.Here the enthalpy value is negative and its means that it the system releases the energy to the surroundings. It is an exothermic process.
Now let’s write an equation which relates enthalpy ($\Delta H$) and the internal energy ($\Delta U$).
And the equation is, \[\Delta H=\Delta U+\Delta {{n}_{g}}RT\]
Here,\[\Delta H=\Delta U+\Delta {{n}_{g}}RT\]
\[\Delta {{n}_{g}}\]= number of moles of gas, T is temperature and R is gas constant
We calculate \[\Delta {{n}_{g}}\]=no. of moles of gas in product side – no. of moles of gas in reactant side
\[\Delta {{n}_{g}}\]=0-1=-1, here in the product side there are no gas molecules whereas there is 1 mole of gas in the reactant side.
Substitute the value of, \[\Delta {{n}_{g}}\] and $\Delta H$ in the above equation relating $\Delta H$ and $\Delta U$
Therefore the equation becomes,
$-3RT=\Delta U+\left( -1 \right)RT=-3RT=\Delta U-2RT$
$-2RT=\Delta U$
-2RT is greater than -3RT, hence\[\Delta U\rangle \Delta H\], change in internal energy is greater than change in enthalpy,
So from the above options the correct answer is option C, which is$\left| \Delta H \right|\langle \left| \Delta U \right|$
Note: $-2RT=\Delta U$
We can use the equation mentioned below to find internal energy parameters, change in number of moles of gas in a reaction, enthalpy.
\[\Delta H=\Delta U+\Delta {{n}_{g}}RT\]
Attention should be given for substituting the value of Since R can have different values in different units.
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