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Assignment Directions: Pick one of the following chemists and perform a bit of research on him/her. Answer the following questio

ns. Alice Hamilton Rosalind Franklin Marie Curie Gertrude B. Elion Ada Yonath Henry Cavendish Robert Boyle Antoine Lavoisier Mario J. Molina Svante Arrhenius

1 answer:

avanturin [10]2 years ago

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Don’t worry about it, let me do your homework!

Born February 27, 1869, Alice Hamilton was an American physician, research scientist, and author who is best known as a leading expert in the field of occupational health and a pioneer in the field of industrial toxicology. She was also the first woman appointed to the faculty of Harvard University.

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Read 2 more answers

Which of the following reactions will have the largest value of K at 298 K? A) CaCO3(s) → CaO(s) + CO2(g) ΔG° =+131.1 kJ B) 2 Hg

Nadya [2.5K]

Answer :

(a) The value of 'K' for the reaction is $1.05\times 10^{-23}$

(b) The value of 'K' for the reaction is $4.93\times 10^{31}$

(c) The value of 'K' for the reaction is $7.17\times 10^{-58}$

(d) The value of 'K' for the reaction is $8.09\times 10^{4}$

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs free energy

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K = equilibrium constant

Now we have to calculate the value of 'K' for the following reactions.

(a) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$ $\Delta G^o=+131.1kJ=+131100J$

$\Delta G^o=-RT\times \ln K$

$+131100J=-(8.314J/K.mol)\times (298K)\times \ln K$

$K=1.05\times 10^{-23}$

Thus, the value of 'K' for the reaction is $1.05\times 10^{-23}$

(b) $2Hg(g)+O_2(g)\rightarrow 2HgO(s)$ $\Delta G^o=-180.8kJ=-180800J$

$\Delta G^o=-RT\times \ln K$

$-180800J=-(8.314J/K.mol)\times (298K)\times \ln K$

$K=4.93\times 10^{31}$

Thus, the value of 'K' for the reaction is $4.93\times 10^{31}$

(c) $3O_2(g)\rightarrow 2O_3(g)$ $\Delta G^o=+326kJ=+326000J$

$\Delta G^o=-RT\times \ln K$

$+326000J=-(8.314J/K.mol)\times (298K)\times \ln K$

$K=7.17\times 10^{-58}$

Thus, the value of 'K' for the reaction is $7.17\times 10^{-58}$

(d) $Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)(s)$ $\Delta G^o=-28.0kJ=-28000J$

$\Delta G^o=-RT\times \ln K$

$-28000J=-(8.314J/K.mol)\times (298K)\times \ln K$

$K=8.09\times 10^{4}$

Thus, the value of 'K' for the reaction is $8.09\times 10^{4}$

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