The equilibrium constant of a reaction is found to fit the expression below, between 300 K and 600 K: lnK=−2.04(1176K/T)+(2.1∑a,b×10^7K^3)/T3. Find ΔrSΘ and ΔriHΘ a. ΔrSΘ = 6.67 J K-1 mol-1 and ΔrHΘ = 7.19 kJ mol-1 b. ΔrSΘ = 21 J K-1 mol-1 and ΔrHΘ = 19.5 kJ mol-1 c. ΔrSΘ = 6.67 J K-1 mol-1 and ΔrHΘ = 19.5 kJ mol-1 d.
The equilibrium constant of a reaction is found to fit the expression below, between 300 K and 600 K: lnK=−2.04(1176K/T)+(2.1∑a,b×10^7K^3)/T3. Find ΔrSΘ and ΔriHΘ a. ΔrSΘ = 6.67 J K-1 mol-1 and ΔrHΘ = 7.19 kJ mol-1 b. ΔrSΘ = 21 J K-1 mol-1 and ΔrHΘ = 19.5 kJ mol-1 c. ΔrSΘ = 6.67 J K-1 mol-1 and ΔrHΘ = 19.5 kJ mol-1 d.
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter17: Chemcial Thermodynamics
Section: Chapter Questions
Problem 17.89QE: For each reaction, an equilibrium constant at 298 K is given. Calculate G for each reaction. (a)...
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The equilibrium constant of a reaction is found to fit the expression below, between 300 K and 600 K:
lnK=−2.04(1176K/T)+(2.1∑a,b×10^7K^3)/T3.
Find ΔrSΘ and ΔriHΘ
a. ΔrSΘ = 6.67 J K-1 mol-1 and ΔrHΘ = 7.19 kJ mol-1
b. ΔrSΘ = 21 J K-1 mol-1 and ΔrHΘ = 19.5 kJ mol-1
c. ΔrSΘ = 6.67 J K-1 mol-1 and ΔrHΘ = 19.5 kJ mol-1
d.
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