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A reaction has ΔrH∘=ΔrH∘= -116 kJmol−1 and ΔrS∘=ΔrS∘= 303 JK−1mol−1 . |
Part A At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings? |
ΔrH∘= -116 kJmol−1 and ΔrS∘= 303 JK−1mol−1 .
S0
=
S surr
S0
= -
H0/T
T =
-
H0/
S0
= -*-116*10^3/303
= 382.84K >>>>>answer
A reaction has ΔrH∘=ΔrH∘= -116 kJmol−1 and ΔrS∘=ΔrS∘= 303 JK−1mol−1 . Part A At what temperature...
Given the values of ΔrH∘, ΔrS∘, and T below, determine ΔSuniv. Part A: ΔrH∘= 130 kJmol−1 , ΔrS∘=− 263 JK−1mol−1 , T= 291 K . Part B ΔrH∘=− 130 kJmol−1 , ΔrS∘= 263 JK−1mol−1 , T= 291 K . Part C ΔrH∘=− 130 kJmol−1 , ΔrS∘=− 263 JK−1mol−1 , T= 291 K . Part D ΔrH∘=− 130 kJmol−1 , ΔrS∘=− 263 JK−1mol−1 , T= 561 K .
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2Ca(s)+O2(g)→2CaO(s)
ΔrH∘= -1269.8 kJmol−1; ΔrS∘= -364.6 JK−1mol−1
Calculate the Gibbs energy change for the reaction at 24 ∘C.
Express your answer using four significant figures.
2Ca(s) + O2 (g) → 2CaO(s) Δ,Ho =-1269.8 kJ mol 1: Δ,S":-3646 J K 1 mol 1 Part A Calculate the Gibbs energy change for the reaction at 24 °C. Express your answer using four significant figures. kJ mol 1 Submit Previous Answers Request Answer
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a reaction has a delta Hrxn= -133 Kj and delta Srxn= 317 J/K.
at what temperature is the change in entropy for the reaction equal
to the change in entropy for the surroundings?
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