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 $m o l^{- 1}$ .

Open in App

Solution

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

${Δ S}_{s u r r .} = \frac{q_{s u r r .}}{T}$

$∴ {Δ S}_{s u r r .} = \frac{286}{298} = 0.96 K J / m o l - K$

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

Suggest Corrections

Similar questions

Calculate the entropy change in surroundings when 1.00 mol of H₂O₍_l₎ is formed under standard conditions. Δ_fH^θ = –286 kJ mol^–1.

Calculate the entropy change in surroundings when 1.00 mol of $H_{2} O_{(l)}$ is formed under standard conditions. $Δ_{f} H^{θ} = - 286 k J m o l^{- 1}$ .

Q. Calculate the entropy change in surrounding when 1.00 mol of

H_{2} O (l)

is formed under standard condition

△_{f} H^{⊖} = - 286 K J m o l^{- 1}

Q. Calculate the entropy change in surroundings when 1.00 mol of water is formed under standard conditions.

Calculate the free energy change when 1 mol of NaCl is dissolved in water at 298 K. given:
(i) Lattice energy of NaCl = -778 kJ $m o l^{- 1}$
(ii) Hydration energy of NaCl = -778 kJ $m o l^{- 1}$
(iii) Entropy change at 298 K = 43 J $m o l^{- 1}$

Join BYJU'S Learning Program

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions at 298 K. Given ΔrH0 = - 286 kJ mol−1.

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 $m o l^{- 1}$ .