Question

For a given reaction, ΔH = -26.6 kJ/mol and ΔS = -77.0 J/K⋅mol. The reaction is spontaneous ________. Assume that ΔH and ΔS do not vary with temperature.

For a given reaction, = -26.6 kJ/mol and = -77.0 J/Kmol. The reaction is spontaneous ________. Assume that and do not vary with temperature.

at T > 298 K | |

at all temperatures | |

at T < 345 K | |

at T < 298 K | |

at T > 345 K |

Answer #1

reaction is spontaneous if delta G is less than zero

The equation for calculating ΔG is the following:

ΔG = ΔH – T(ΔS)

Gibbs Free Energy and must be negative for a spontaneous reaction. Set delta G = 0 (for equilibrium) and solve for T

ΔH = -26.6 kJ/mol = -26600 j/mol

ΔS = -77.0 J/K⋅mol

ΔG = -26600 -T(-77.0)

0 = -26600 +(77.0)T

26600= 77.0T

T = 26600/77

T= 345

delta S is negative, so (-T delta S) is positive and G is more
positive for higher temperatures

This reaction is spontaneous for temperatures less than 345 K

**so .at T < 345 K reaction is spontaneous**

A reaction for which ΔH° = + 98.8 kJ and ΔS° = + 141.5 J/K is
______________________(spontaneous or nonspontaneous) at low
temperatures and________________________ (spontaneous or
nonspontaneous) at high temperatures.

Given the values of ΔH∘rxn, ΔS∘rxn, and
T below, determine ΔSuniv.
Part A
ΔH∘rxn= 115 kJ , ΔS∘rxn=− 263 J/K ,
T= 301 K .
ΔSuniv=
J/K
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Part B
ΔH∘rxn=− 115 kJ , ΔS∘rxn= 263 J/K ,
T= 301 K .
ΔSuniv=
J/K
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Part C
ΔH∘rxn=− 115 kJ , ΔS∘rxn=− 263 J/K ,
T= 301 K .
ΔSuniv=
J/K
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Part D
ΔH∘rxn=− 115 kJ , ΔS∘rxn=− 263 J/K ,
T= 557...

Given the values of ΔH∘rxn, ΔS∘rxn, and
T below, determine ΔSuniv.
Part A
ΔH∘rxn= 117 kJ , ΔS∘rxn=− 263 J/K ,
T= 291 K .
ΔSuniv=
−1.3•102
J/K
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Part B
ΔH∘rxn=− 117 kJ , ΔS∘rxn= 263 J/K ,
T= 291 K .
ΔSuniv=
J/K
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Part C
ΔH∘rxn=− 117 kJ , ΔS∘rxn=− 263 J/K ,
T= 291 K.
ΔSuniv=
J/K
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Part D
ΔH∘rxn=− 117 kJ ,...

For the reaction
2C2H6(g) +
7O2(g) ---->
4CO2(g) +
6H2O(g)
ΔH° = -2855.4 kJ and ΔS° = 92.7
J/K
The maximum amount of work that could be done when
2.39 moles of
C2H6(g) react at
349 K, 1 atm is ____ kJ.
Assume that ΔH° and ΔS° are independent of temperature.

From the values of ΔΗ and ΔS, predict which of the following
reactions would be spontaneous at 25oC. If the reaction
is nonspontaneous at 25oC, at what temperature would it
become spontaneous?
A) ΔH = 10.5 kJ/mol, ΔS = 30 J/K * mol
B) ΔΗ = 1.8 kJ/mol, ΔS= -113 J/K * mol

1- Above what temperature is the following reaction
spontaneous?
N2O4(g) ↔ 2
NO2(g)
ΔH° = 57.24 kJ/mol ΔS° = 175.5 J/mol∙K
Group of answer choices
326 K
53.2 K
307 K
273 K
2- Predict the sign on ΔG for the following reaction when
PI2 = PH2 = 0.01 atm and PHI = 1.0
atm.
H2(g) + I2(g) ↔ 2
HI(g) ΔG° = -15.94 kJ/mol & Kp,298K =
620
Group of answer choices
ΔG = 0
ΔG > 0
ΔG...

The
enthalpy of a reaction is 50 kj/mol, and the enthaply is 223 J/K
mol. What is the minimum temperature for this reaction to be
spontaneous?

For a particular reaction, delta H is 67.6 kJ/mol and
delta S is 126.9 J/(molk). Assuming these values change very little
with temperature, over what temperature range is the reaction
spontaneous in the forward direction?
The reaction is spontaneous for temperatures
less than? Greater than?
T= ???? K

Given that ΔH∘=−92.38 kJ and ΔS∘=−198.3 J/K, what is the
temperature above which the Haber ammonia process becomes
nonspontaneous? Given that and , what is the temperature above
which the Haber ammonia process becomes nonspontaneous?
25 ∘C
47 ∘C
61 ∘C
193 ∘C
500 ∘C

For the gaseous reaction of xenon (Xe) and fluorine (F) to form
xenon hexafluoride (XeF6), ΔH° = -402 kJ/mol and ΔG° =
-280 kJ/mol at 298 K.
Xe (g) + 3F2 (g) → XeF6 (g)
a. Calculate the ΔS° for the reaction.
b. Calculate the ΔG° at 500 K.
c. Calculate the Keq values for the reaction at 298 K
and 500 K.
d. At what temperature value does the reaction become
non-spontaneous?

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