Question

For the hypothetical reaction:
${{A}_{2}}(g)+{{B}_{2}}(g)\rightleftharpoons 2AB(g)$
${{\Delta }_{r}}{{G}^{o}}$ and ${{\Delta }_{r}}{{S}^{o}}$ is 20 kJ/mol and $-20J{{K}^{-1}}mo{{l}^{-1}}$respectively at 200 K. If ${{\Delta }_{r}}{{C}_{p}}$ is $20J{{K}^{-1}}mo{{l}^{-1}}$, then${{\Delta }_{r}}{{H}^{o}}$ at 400 K is:
A. 20 kJ/ mol
B. 7.98 kJ/ mol
C. 28 kJ/ mol
D. none of these

Answer 1

The standard state of any reactant substance is denoted by r in the subscript, here ${{\Delta }_{r}}{{G}^{o}}$ and ${{\Delta }_{r}}{{S}^{o}}$ are the free energy and entropy in the standard state respectively. Change in enthalpy ${{\Delta }_{r}}{{H}^{o}}$ is the sum of change in energy and specific heat capacity at constant pressure ${{\Delta }_{r}}{{C}_{p}}$ multiplies by change in time.

Complete answer:
We have been given a hypothetical reaction, where all the reactants are in standard forms, which are gases. We have given free energy and entropy change at 200 K temperature as 20 kJ/mol and $-20J{{K}^{-1}}mo{{l}^{-1}}$ respectively. Also the specific heat capacity at constant pressure, ${{\Delta }_{r}}{{C}_{p}}$ is given to be $20J{{K}^{-1}}mo{{l}^{-1}}$. We have to find the enthalpy change at 400 K temperature.
From the relation of free energy, entropy and enthalpy change we have,
$\Delta {{H}_{1}}=\Delta {{G}^{0}}+T\Delta {{S}^{0}}$
Where ${{\Delta }_{r}}{{G}^{o}}$ and ${{\Delta }_{r}}{{S}^{o}}$ is 20 kJ/mol and $-20J{{K}^{-1}}mo{{l}^{-1}}$ respectively and T is 200 K, so
$\Delta {{H}_{1}}=20+200(-20)$
$\Delta {{H}_{1}}=-3980K\,J$
Now, from the relation between, enthalpy change, energy change, and heat capacity at constant pressure, we will take out the value of ${{\Delta }_{r}}{{H}^{o}}$ at 400 K. as the energy change$\Delta {{H}_{1}}=-3980K\,J$ is the $\Delta E$.
So, we have $\Delta {{H}^{0}}=\Delta E+{{C}_{p}}\Delta T$, given ${{\Delta }_{r}}{{C}_{p}}$ is $20J{{K}^{-1}}mo{{l}^{-1}}$,and T is 400 K and 200 K.
$\Delta {{H}^{0}}=-3980+20(400-200)$
$\Delta {{H}^{0}}=-3980+4000$
$\Delta {{H}^{0}}=20kJ/mole$
Therefore ${{\Delta }_{r}}{{H}^{o}}$is 20 kJ/ mole.

so option A is correct.

Note:
Enthalpy change and energy change are considered as same values, as $\Delta E=\Delta H$, first we have taken out the enthalpy by putting the entropy and free energy values, then this enthalpy is used as the internal energy for calculating the standard enthalpy at specific heat at constant pressure with the given time range.