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Zarrin [17]
3 years ago
9

Imagine the reaction A + B LaTeX: \Longleftrightarrow⟺ C + D proceeds at room temperature (25 °C) and is determined to have a re

action 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.
Engineering
1 answer:
Wittaler [7]3 years ago
5 0

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.

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