An ideal gas undergoes a four step cycle as shown in the P − V diagram below. During this cycle, heat is absorbed by the gas in
($\alpha$) steps 1 and 2 ($\beta$) steps 1 and 3
($\gamma$) steps 1 and 4 ($\delta$) steps 2 and 4 Continue reading An ideal gas undergoes a four step cycle …
For the reaction $A(g) \rightleftharpoons B(g)$ at 495 K, $\Delta_r G^\circ = -9.478 kJ.mol^{-1} $. If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is _ _ _ _ millimoles. (Round off to the nearest integer)
[$R=8.314 J.mol^{-1} .K^{-1} $; ln 10 = 2.303 ] Continue reading For the reaction $A(g) \rightleftharpoons B(g)$ at 495 K ….
For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g) $, when $\Delta S = -176.0 JK^{-1} $ and $\Delta H = -57.8 kJ mol^{-1} $, the magnitude of $\Delta G $ at 298 K for the reaction is _ _ _ _ $kJ mol^{-1} $. (Nearest integer)
Solution
We have, $\Delta G = \Delta H – T \Delta S $
$\Delta S = -176.0 JK^{-1} = -0.176 kJ K^{-1} $
$\Delta G = -57.8 – 298 \times (-0.176) \approx -57.8+52.5 = -5.3 $
$\therefore |\Delta G | = 5.3 kJ mol^{-1} $
Ans: 5
The word, “Free” in Gibbs Free Energy can be often confusing and I will try to address this aspect in simplified manner. Enthalpy is the total heat content, which means the internal energy + the product of pressure and volume. However, all enthalpy is not available to do work. The reason is the entropy or disorder. Nature seems to prefer disorder in broader sense as it does not like to spend energy to arrange things (there may be symmetries and orderliness in local/system/surrounding context but in broader context, i.e. in the context of the universe/multiverse, the disorder wins). Some enthalpy is spent in entropy. The remaining is left to do work and because that is what is available to do work, it is called available energy or free energy.
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