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Select the most apropriate option · Q7

Q.vii. If the standard enthalpy of formation of methanol is -238.9 kJ mol−1^{-1} then entropy change of the surroundings will be a. -801.7 J K−1^{-1}
b. 801.7 J K−1^{-1}
c. 0.8017 J K−1^{-1}
d. -0.8017 J K−1^{-1}

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Step 1. ΔSsurr = -ΔH/T, where ΔH is the SYSTEM's enthalpy change (here the formation enthalpy of methanol, -238.9 kJ/mol = -238900 J/mol).

Step 2. ΔSsurr = -(-238900 J)/298 K = +801.7 J K-1. …

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