Question

Create the equation of a line by substituting delta G=delta H-TdeltaS into the equation delta G= -RT ln(Kc), and solve for ln(Kc)

Answer #1

ΔG = -RTlnKc -----(1)

and

ΔG = ΔH – TΔS -----(2)

The Equation 1 and 2 can be combined to obtain the Equation 3 as follows:

-RTlnKc = ΔH – TΔS ------- (3)

Reorganization of Equation 3 yields Equation 4, which shows the linear relationship between lnKsp and 1/T:

lnKc = − ( ΔH/R ) ( 1/T ) + ΔS/R ----------(4)

Therefore, a plot of lnKc versus 1/T yields a straight line with a slope of –ΔH/R and a y-intercept of ΔS/R. So the ΔH and ΔS can be calculated based on the linear equation obtained from the linear fitting of the graph.

CH4(g)+4Cl2(g) —>CCl4(l) + 4HCl(g)
2.calculate the standard molar enthalpy (delta H ) of the
reaction
3.Explain if the reaction/process is always spontaneous,never
spontaneous,or spontaneous only at high temperatures.(Hint:
consider how delta H and deltaS influence delta G and spontaneity
with the equation delta G =delta H -T deltaS )

Rewrite equation with the delta H value included in either
reactants or products and identify the reaction as either endo- or
exo- thermic: 2 NO2 (g) --> 2 NO (g) + O2
(g) H= 114.2 kJ/mol

1)When delta G= 0; a system is at equilibrium. Use values for
delta H and delta S to determine the temperature at which the
following system would reach equilibrium.
N2(g)+ O2(g)--->2NO(g)
2)Complete te following table:
Enthalpy change delta H, Etropy change delta S, Predicted Sign
for delta G.
+(positive) -(negatve) ?
+(positive) -(negative) ?
-(negative) -(negative) ?
-(negative) +(positive) ?

Use
gibb's free energy equation to calculate the delta G of a reaction
with the following values: 298.15 K, -.1434 kJ/molK (delta S),
-1104.14 kJ (delta H).

Calculate delta G at 82 C for reactions in which
a. delta H = 293 kj delta S = -695 J/K
b. delta H = -1137 kJ; delta S = 0.496 kJ/K
c. delta H = -86.6 kJ; delta S = -382 J/K

calculate the standard free energy change delta G for reaction
N2 (g) +3H2(g)—>2NH3
N2 delta H=0.00kj mol^-1s=+191.5J mol^-1K^-1
H2 delta H=0.00kj mol^-1,s = +130.6j mol^-1 k-1
NH3 delta H=-46.0kj mol^-1,s =192.5 J mol^-1 k-1
A. +112.3 kJ
B.-87.6kJ
C.-7.4kJ
D.-32.9 kJ
E.-151.1kJ

You are given the following data:
H2(g) --> 2H(g) Delta H degrees=436.4 kJ/mol
Br2(g) --> 2Br(g) Delta H degrees=192.5 kJ/mol
H2(g) + Br2(g) --> 2HBr(g) Delta H
degrees=-72.4 kJ/mol
Calculate Delta H degrees for the reaction
H(g) + Br(g) --> HBr(g)

Apply the Free Energy equation to identify whether reactions
with the following values of delta H and Delta S are spontaneous or
nonspontaneous and whether they are exothermic or endothermic:
a. delta H = -128 kJ; delta S = 35 J/K at 500k
b. delta H = +67kJ; delta S= -140 J/K at 250k
c. delta H = +75 kJ; delta S = 95J/K at 800k

Q1Calculate the delta H for N2 (g) + O2 (g)------>2NO(g)
4NH3(g) +5O2(g)--->4NO(g)+6H2O(l). DeltaH= -1170kJ
4NH3(g) +
3O2(g)--->2N2(g)+6H2O(l) Delta H =-1530kJ
Q2Calculate standard heat of
formation delta H degree f. For NCl 3
NH3(g)+3HCl
(g)--->NCl3(g)+3H2(g). Delta H =564.8kJ

Calculate the delta H degrees reaction for the following
reaction:
2Ni(s) + 2S(s) +3O2 (g) -> 2NiSO3(s)
from the following info:
NiSO3(s) -> NiO(s) + SO2(g) delta H=156kJ
S(s) + O2(g) -> SO2(g) delta H= -297
Ni(s) + 1/2O2(g) -> NiO(s) delta H = -241

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