At 25 ∘c25 ∘c, the equilibrium partial pressures for the reaction 3a(g)+2b(g)↽−−⇀c(g)+2d(g) 3a(g)+2b(g)↽−−⇀c(g)+2d(g) were found to be pa=5.84pa=5.84 atm, pb=4.47pb=4.47 atm, pc=4.17pc=4.17 atm, and pd=4.32pd=4.32 atm. what is the standard change in gibbs free energy of this reaction at 25 ∘c25 ∘c ?

9742.37 J/mol

Explanation:

We follow the expression for Gibbs free energy:

First we calculate Q with the following expression for the reaction:

Therefore, Q=0.0196

Now we continue with the first equation, and as we are on equilibrium, we know that the Gibbs free energy is zero, therefore:

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explanation:

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the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.