Calculate the entropy change in surroundings when 1.00 mol of $$H_2O$$(l) is formed under standard conditions. $$Δ_f H^ Θ= –286 kJ mol^{–1}.$$

Asked by Abhisek | 1 year ago |  132

Solution :-

It is given that $$−286\;kJmol^{−1}​​​​​​​$$of heat is evolved on the formation of 1 mol of H2O(l) . Thus, an equal amount of heat will be absorbed by the surroundings.

Thus, the same heat will be absorbed by surrounding $$q_{surr}$$

$$+286\;kJmol^{-1}.$$

Now, $$\Delta S_{surr} = \dfrac{q_{surr}}{7}$$

$$\dfrac{286kJmol^{-1}}{298K}$$

Therefore, $$\Delta S_{surr} = 959.73Jmol^{-1}K^{-1}$$

Answered by Pragya Singh | 1 year ago

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