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If a reaction has ΔH = 100 kJ and ΔS = -200 J/K, what is ΔG at 298 K?
If a reaction has ΔH = 100 kJ and ΔS = -200 J/K, what is ΔG at 298 K?
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If a reaction has ΔH = 100 kJ and ΔS = -200 J/K, what is ΔG at 298 K?
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ΔG = ΔH - TΔS = 100 kJ - 298 K * (-0.2 kJ/K) = 100 kJ + 59.6 kJ = 159.6 kJ.
Questions & Step-by-step Solutions
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Q: If a reaction has ΔH = 100 kJ and ΔS = -200 J/K, what is ΔG at 298 K?
Solution:
ΔG = ΔH - TΔS = 100 kJ - 298 K * (-0.2 kJ/K) = 100 kJ + 59.6 kJ = 159.6 kJ.
Steps: 8
Show Steps
Step 1: Identify the values given in the problem. We have ΔH = 100 kJ and ΔS = -200 J/K.
Step 2: Convert ΔS from J/K to kJ/K because ΔH is in kJ. Since 1 kJ = 1000 J, we convert -200 J/K to kJ/K: -200 J/K = -0.2 kJ/K.
Step 3: Identify the temperature (T) given in the problem, which is 298 K.
Step 4: Use the formula for Gibbs free energy: ΔG = ΔH - TΔS.
Step 5: Substitute the values into the formula: ΔG = 100 kJ - 298 K * (-0.2 kJ/K).
Step 6: Calculate TΔS: 298 K * (-0.2 kJ/K) = -59.6 kJ.
Step 7: Now substitute this value back into the equation: ΔG = 100 kJ + 59.6 kJ.
Step 8: Finally, calculate ΔG: ΔG = 159.6 kJ.
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