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

Answer:

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

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

$\Delta G_{rxn}=-RT\ln K+RT\ln Q$

$=-1.987 cal/K mol\times 298 K\ln [35]+1.987 cal/K mol\times 298K\times \ln [46]$

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

1 cal = 0.001 kcal

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