Question

Imagine the reaction A + B LaTeX: \Longleftrightarrow⟺ C + D proceeds at room temperature (25 °C) and is determined to have a reaction quotient (Q) equal to 46. If the equilibrium constant (Keq) for this reaction is equal to 35, what is the value of LaTeX: \DeltaΔG for this reaction in kcal/mol. Report your answer to the nearest tenth. Recall that to convert from °C to K you simply add 273 to the temperature in °C. In addition, the gas constant R is equal to 1.987 cal/KLaTeX: \cdot⋅mol. Be careful of units.

Answer 1

0.2 kcal/mol is the value of
`\Delta G` for this reaction.

Step-by-step explanation:

The formula used for is:

`\Delta G_(rxn)=\Delta G^o+RT\ln Q`

`\Delta G^o=-RT\ln K`

where,

`\Delta G_(rxn)` = Gibbs free energy for the reaction

`\Delta G_^o` = standard Gibbs free energy

R =Universal gas constant

T = temperature

Q = reaction quotient

k = Equilibrium constant

We have :

Reaction quotient of the reaction = Q = 46

Equilibrium constant of reaction = K = 35

Temperature of reaction = T = 25°C = 25 + 273 K = 298 K

R = 1.987 cal/K mol

`\Delta G_(rxn)=-RT\ln K+RT\ln Q`

`=-1.987 cal/K mol* 298 K\ln [35]+1.987 cal/K mol* 298K* \ln [46]`

`=-2,105.21 cal/mol+2,267.04 cal/mol=161.82 cal/mol=0.16182 kcal/mol\approx 0.2 kcal/mol`

1 cal = 0.001 kcal

0.2 kcal/mol is the value of
`\Delta G` for this reaction.