Question

Question 4 1 pts Use the standard free energies of formation (AG) of each substance to determine the value of equilibrium constant K at 25°C for the following reaction. CO2(g) + 2 H2(g) = CH2OH() AGE® kJ/mol -394.4 0 -166.4 8.83 * 1039 1.13 x 10-40 00 0.91 Question 5 Select ALL that are true for a chemical reaction at equilibrium. AG=0 rate of forward reaction equals the rate of reverse reactions O Q=K OK=1 The reaction has stopped.

Answer 1

We know that for any reaction:-

aA+bB→cC+dD

ΔrGo=cΔfGo(C)+dΔfGo(D)−aΔfGo(A)−bΔfGo(B)

ΔfG0=∑vΔfG0(products)−∑vΔfG0(reactants)

ΔfG0 = -164.4 -(-394.4)

ΔfG0 = 230 KJ/mol

As we know for a equillibrium reaction ΔfG = 0 KJ/mol

then the expression become-

ΔGo=−RTlnKeq
230000=−8.314 * 298 * lnKeqlnKeq = -92.83282 and if we round off the value to 92 for our convineance:-

We get Keq = 1.1 * 10^-40 So, option 2 will be correct for this question.

5. For a chemical reaction at equillibrium :- The true statements are:-o

ΔG = 0

The rate of forward reaction = rate of backward reaction

Q = K

Yes, K is about equal to 1, the reaction will reach equilibrium.