What is ΔrH for a reaction that has ΔrG = -0.2 kJ mol-1 and ΔrS =...

Question

Question

What is Δ_rH for a reaction that has Δ_rG = -0.2 kJ mol^-1 and Δ_rS = 65.4 J mol^-1 K^-1 at 48.2 °C?

Express your answer in kJ mol^-1 .

Answer 1

Answer #1

ΔGo = -0.2 KJ/mol

ΔSo = 65.4 J/mol.K

= 0.0654 KJ/mol.K

T= 48.2 oC

= (48.2+273) K

= 321.2 K

use:

ΔGo = ΔHo - T*ΔSo

-0.2 = ΔHo - 321.2 * 0.0654

ΔHo = 20.8 KJ/mol

Answer: 20.8 KJ/mol

Answer 2

Similar Homework Help Questions

What is ΔrH for a reaction that has ΔrG = -26.3 kJ mol-1 and ΔrS =...

What is ΔrH for a reaction that has ΔrG = -26.3 kJ mol-1 and ΔrS = 44.9 J mol-1 K-1 at -9.0 °C? Express your answer in kJ mol-1.
What is ΔrG for a reaction that has ΔrH = 86.7 kJ mol-1 and ΔrS =...

What is ΔrG for a reaction that has ΔrH = 86.7 kJ mol-1 and ΔrS = 83.7 J mol-1 K-1 at 3.3 °C? Express your answer in kJ mol-1 .
What is ΔrS for a reaction that has ΔrH = -85.6 kJ mol-1 and ΔrG =...

What is ΔrS for a reaction that has ΔrH = -85.6 kJ mol-1 and ΔrG = 93.9 KJ mol-1 at 16.2 °C?

What is ΔrSfor a reaction that has ΔrH = -50.1 kJ mol-1 and ΔrG = -77.9...

What is ΔrSfor a reaction that has ΔrH = -50.1 kJ mol-1 and ΔrG = -77.9 KJ mol-1 at 4.9 °C? Express your answer in J mol-1 K-1.
For the decomposition of barium carbonate, consider the following thermodynamic data: ΔrH∘ΔrH∘ 271.5kJ mol−1 ΔrS∘ΔrS∘ 173.8J...

For the decomposition of barium carbonate, consider the following thermodynamic data: ΔrH∘ΔrH∘ 271.5kJ mol−1 ΔrS∘ΔrS∘ 173.8J K−1 mol−1 A: Calculate the temperature in kelvins above which this reaction is spontaneous. Answer:1562K B: Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C: BaCO3(s)→BaO(s)+CO2(g) Answer: 2.73*10^-39 C: When adjusted for any changes in ΔrH and ΔrS with temperature, ΔrG∘(600K)=167kJ mol−1. Calculate the equilibrium constant at 600 K .
A reaction has ΔrH∘=ΔrH∘= -116 kJmol−1 and ΔrS∘=ΔrS∘= 303 JK−1mol−1 . Part A At what temperature...

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?

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...

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
1--What is the ΔrG for the following reaction (in kJ mol-1) at 298 K? CH4 (g)...

1--What is the ΔrG for the following reaction (in kJ mol-1) at 298 K? CH4 (g) + 2 O2 (g) ⇌ CO2 (g) + 2 H2O (g) The conditions for this reaction are: PCH4 = 0.73 bar PO2 = 0.27 bar PCO2 = 0.10 bar PH2O = 1.41 bar 2--What is the ΔrG for the following reaction (in kJ mol-1) at 298 K? 2 Si (s) + 3 H2 (g) ⇌ Si2H6(g) The conditions for this reaction are: PH2 =...
What is ΔrG° for a reaction (in kJ mol-1) which has equilibrium constant K = 1.0...

What is ΔrG° for a reaction (in kJ mol-1) which has equilibrium constant K = 1.0 x 1010 at the temperature 16.3 °C? You have 5 attempts at this question. Remember: if you want to express an answer in scientific notation, use the letter "E". For example "4.32 x 104" should be entered as "4.32E4".

Consider the following reaction: 4??2(?) +?2(?) +2?2?(?) →4???3(?) ΔrG° = -70.8 kJ mol-1 at 298 K,...

Consider the following reaction: 4??2(?) +?2(?) +2?2?(?) →4???3(?) ΔrG° = -70.8 kJ mol-1 at 298 K, and ΔrH° = -346 kJ mol-1 a) What is the equilibrium constant at 298 K? b) At what temperature (in K) will K = 1? just need help with B PLEASE