Question

The reaction \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) has an equilibrium constant of \(5.8\) and under certain conditions has \(Q=336\)

Part \(A\)

What can you conclude about the sign of \(\Delta G_{\mathrm{xxn}}^{a}\) and \(\Delta G_{\mathrm{rxa}}\) for this reaction under these conditions?

Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

Answer 1

Chapter 18 Homework Exercise 18.116 30 of 37 > Review Constants Periodic Table Part A The reaction A (g) = B(9) has an equilibrium constant of 5.8 and under certain conditions has Q=336. What can you conclude about the sign of AG. and AG xn for this reaction under these conditions? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Reset Help The relationship between AG and I AGRx = -RTINK en K = 5.8, the natural log of K is positive and AG is less than 0. When Q = 336, the second term in AG +RTIQ is positive so at high temperature, this term can dominate and make AG greater than 0. AG AG7x = AGA + RTIQ Submit Request Answer

Since the equilibrium constant is greater than 1, the value of In K. is a positive value. Substituting the all the values in the formula for AGºran gives a negative value.

The relation between free energy and reaction quotient is as follows: AGF = AGºren+RT In Q But, it is known that, AGºran =-RT In K. so, AG = (-RT In K.)+(RT InQ) = RT (In Q-In K.) Given, K = 5.8 Q=336 Substitute the values as follows: AG H = RT (In Q- In K.) = RT (In 336 - In 5.8) = RT (4.05) Substituting the all the values in the formula for AGran gives a positive value. So, AGrx >0.