Chemistry, 23.04.2021 07:10 YoVeoAnime
An ice cube at 0°C melts when placed inside a room at 22°C (295K). Based on the concepts of enthalpy, entropy, and Gibbs free energy, which of the following best explains why the process is thermodynamically favorable?
Melting ice releases energy, ΔH°<0
, and ΔS°<0
because the motion of the H2O
molecules decreases as it transitions from solid to liquid. At a temperature higher than 0°C
(273K)
, the term TΔS°
is smaller than ΔH°
, resulting in a thermodynamically favorable process with ΔG°<0
.
Melting ice releases energy, delta H naught is less than 0 , and delta S naught is less than 0 because the motion of the H 2 O molecules decreases as it transitions from solid to liquid. At a temperature higher than 0 degrees Celsius or 273 kelvins , the term T delta S naught is smaller than delta H naught , resulting in a thermodynamically favorable process with delta G naught is less than 0 .
A
Melting ice releases energy, ΔH°<0
, and ΔS°>0
because the motion of the H2O
molecules increases as it transitions from solid to liquid. At a temperature higher than 0°C
(273K)
, the term TΔS°
is greater than ΔH°
, resulting in a thermodynamically favorable process with ΔG°>0
.
Melting ice releases energy, delta H naught is less than 0 , and delta S naught is greater than 0 because the motion of the H 2 O molecules increases as it transitions from solid to liquid. At a temperature higher than 0 degrees Celsius or 273 kelvins , the term T delta S naught is greater than delta H naught , resulting in a thermodynamically favorable process with delta G naught is greater than 0 .
B
Melting ice requires energy, ΔH°>0
, and ΔS°<0
because the motion of the H2O
molecules decreases as it transitions from solid to liquid. At a temperature higher than 0°C
(273K)
, the term TΔS°
is smaller than ΔH°
, resulting in a thermodynamically favorable process with ΔG°>0
.
Melting ice requires energy, delta H naught is greater than 0 , and delta S naught is less than 0 because the motion of the H 2 O molecules decreases as it transitions from solid to liquid. At a temperature higher than 0 degrees Celsius or 273 kelvins , the term T delta S naught is smaller than delta H naught , resulting in a thermodynamically favorable process with delta G naught is greater than 0 .
C
Melting ice requires energy, ΔH°>0
, and ΔS°>0
because the motion of the H2O
molecules increases as it transitions from solid to liquid. At a temperature higher than 0°C
(273K)
, the term TΔS°
is greater than ΔH°
, resulting in a thermodynamically favorable process with ΔG°<0
.
Answers: 3
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