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

Open in App

Solution

Formation of water is an exothermic process So, $H = - 286 k J / m o l$
Hence heat is transformed to the surroundings
Thus $Q_{s r r} = + 286 k J / m o l$
Now $S = q / t$
$= (286 \times 1000) / 298$
$= 959 J K^{- 1} m o l^{- 1}$

Suggest Corrections

Similar questions

Q. 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}

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}

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

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.

Q. Calculate the entropy change when 1 mol of ice

0^{\circ} C

is converted into water at

0^{\circ} C

. The change in heat is 1436 cal per mol.

Join BYJU'S Learning Program

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

Formation of water is an exothermic process So, H=−286kJ/molHence heat is transformed to the surroundings Thus Qsrr=+286kJ/molNow S=q/t=(286×1000)/298=959JK−1mol−1

Formation of water is an exothermic process So, $H = - 286 k J / m o l$
Hence heat is transformed to the surroundings
Thus $Q_{s r r} = + 286 k J / m o l$
Now $S = q / t$
$= (286 \times 1000) / 298$
$= 959 J K^{- 1} m o l^{- 1}$