1. Tardigrade
  2. Question
  3. Chemistry
  4. If the reaction achieves equilibrium at 298 K one bar pressure i.e., standard conditions then equilibrium constant is:

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Q. If the reaction achieves equilibrium at $298 K \&$ one bar pressure i.e., standard conditions then equilibrium constant is:

Thermodynamics

Solution:

At eqm. $\Delta G=0$

Since standard conditions prevail at eqm,

then $\Delta G=\Delta^{\circ} G =0$

Now, $\Delta^{\circ} G =-2.303 RT \log K$

$0 =-2.303 RT \log K$

$\log K =0$

$K =1$