Question

The reaction a(g)⇌b(g) has an equilibrium constant of 5.8 and under certain conditions has q = 336. part a what can you conclude about the sign of δg∘rxn and δgrxn for this reaction under these conditions?

Answer 1

Answer:

The answer is "As $Q=336$, at high-temperature $\Delta G_{rxn}>0$ and When$K>1,$$\Delta G^{\circ}_{rxn}>0$."

Explanation:

The equation for the reaction is:

$A(g) \leftrightharpoons B(g)$  

$K=5.8\\\\Q=336$

At equilibrium,

$\Delta G^{\circ}_{rxn}>0$$=-RT \ln \ K$

When k=5.8(>1), the value of $\ln k$ would be positive

So, $\Delta G^{\circ}_{rxn}$ is negative (< 0)

So if K > l, $\Delta G^{\circ}_{rxn}<0$

If the reaction is not in equilibrium so the equation is :

$\Delta G_{rxn}>0$=$\Delta G^{\circ}_{rxn}$$+RT \ln Q$

Substituting the expression:

$\Delta G_{rxn}>0$$= (-RT \ln K) + RT \ln Q$

                 $= RT(\ln Q- \ln K)\\= RT(\ln (336)-\ln (5.8))\\= RT(4.06)$

It is the positive value for all temperatures.

So, As Q = 336, at the high temperature $\Delta G_{rxn}>0$.