Question

Consider the following reaction: 2NO(g)+O2(g)⇌2NO2(g) The data in the table show the equilibrium constant for this reac...

Consider the following reaction: 2NO(g)+O2(g)⇌2NO2(g) The data in the table show the equilibrium constant for this reaction measured at several different temperatures.

Temperatures: Kp:

170K 3.8*10^-3

180K 0.34

190K 18.4

200K 681

Part A Use the data to find ΔH∘rxn and ΔS∘rxn for the reaction.

i found ΔH∘rxn = 114 kJ but i dont know how to get  ΔS∘rxn. please help and show work

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Answer #1
Concepts and reason

Plot a graph between on x-axis and on y-axis. A straight line is obtained. Get the equation for the straight line obtained. Find the slope and intercept of the straight line.

Compare the straight line obtained with the equation In K. = 44
. Slope of the straight line is equal to -
H° | R
and intercept is equal to AS / R
.

Fundamentals

Relation between equilibrium constant and standard Gibbs free energy is AGⓇ =-RT In K
. Relation between standard Gibbs free energy change, standard enthalpy change, and standard entropy change is ᎪG° =H° -TAS
. From the two relations: -RT In K, =AH° - TAS
. Here, is gas constant and is temperature in Kelvin scale.

From the given values of temperature () and equilibrium constant (), a table is prepared with and values, as follows:

T(K) 1/1 (K)
170 5.88x10
180 5.56x10°
190 5.26x10-
200 5.00x10
K, In K,
3.80x10- | -5.57
0.34 -1.08
18.4 2.91
6 81 6.52

The graph between on x-axis and on y-axis is as follows:

Slope of the straight line is -13704
and intercept of the straight line is 75.042
.

AH
-= slope
ΔΗ
8.314 JK-mol-=-13704
AFT° = (13704)(8.314 JK-mol)( 10
= 114 kJ mol-1

= intercept
45°
8.314 JK molt=75.042
AS™ = (75.042)(8.314 JK-mol)
AS = 624. JK-mol-

Ans:

The value of is 114 kJ mol
and the value of is 624. JK mol
.

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