Calculate the entropy change in surroundings when 1.00 mol of $H_{2} O (l)$ is formed under standard conditions at 298 K. Given $Δ_{r} H^{0}$ = - 286 kJ $m o l^{- 1}$ .

Question

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Answer 1

$H_{2}_{(g)} + \frac{1}{2} O_{2}_{(g)} ⟶ H_{2} O_{(l)} Δ_{f} H^{0} = - 286 K J / m o l$
From the above equation,

At $298 K$ , when $1$ mole of $H_{2} O_{(l)}$ is formed, $286 K J$ of heat is released. The same amount of heat is absorbed by the surroundings.

$∴ q_{s u r r .} = + 286 K J / m o l; T = 298 K$

As we know that,

$Δ S_{s u r r .} = \frac{q _{s u r r .}}{T}$

$∴ Δ S_{s u r r .} = \frac{2 8 6}{2 9 8} = 0.96 K J / m o l - K$

Hence the entropy change in surroundings will be $0.96 K J / m o l$ .

Answer 2

H_{2}_{(g)} + \frac{1}{2} O_{2}_{(g)} ⟶ H_{2} O_{(l)} Δ_{f} H^{0} = - 286 K J / m o l

From the above equation,

At

298 K

, when

1

mole of

H_{2} O_{(l)}

is formed,

286 K J

of heat is released. The same amount of heat is absorbed by the surroundings.

∴ q_{s u r r .} = + 286 K J / m o l; T = 298 K

As we know that,

Δ S_{s u r r .} = \frac{q _{s u r r .}}{T}

∴ Δ S_{s u r r .} = \frac{2 8 6}{2 9 8} = 0.96 K J / m o l - K

Hence the entropy change in surroundings will be

0.96 K J / m o l

.