Calculate the entropy change at 373 K for the following transformation- H 2 O I- 1-01325 bar - H 2 O g- 0-101325 bar - Given - vap H H 2 O -37-3kJ mol -1

Calculate the change in entropy for the following reaction
$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ .
Given $S_{C O_{2}}^{0} = 213.6 J K^{- 1} m o l^{- 1}$ , $S_{C O}^{0} = 297.6 J K^{- 1} m o l^{- 1}$ respectively.

Q. Calculate the entropy

(J K^{- 1}

mol

^{- 1})

change for the following reversible process:

α - T i n ⇌ β - T i n

13^{\circ} C

1 mol at 1 atm

Δ H_{trans} = 2288 J

mol

^{- 1}

Q. Calculate the entropy change for the following reaction,

H_{2} (g) + C l_{2} (g) ⟶ 2 H C l (g)

298 K

. Given that,

S_{H_{2}}^{⊖} = 131 J K^{- 1} m o l^{- 1}

S_{C l_{2}}^{⊖} = 223 J K^{- 1} m o l^{- 1}

and

S_{H C l}^{⊖} = 187 J K^{- 1} m o l^{- 1}

Q. The molar enthalpy change for

H_{2} O (l) ⇌ H_{2} O (g)

373

K and

1

atm is

41 k J / m o l

. Assuming ideal behavior, the internal energy change for vaporization of

1

mol of water at

373 K

and

1

atm in kJ mol

^{- 1}

is:

Join BYJU'S Learning Program

Calculate the entropy change at 373 K for the following transformation.H2O(I,1.01325bar)→H2O(g,0.101325bar). Given ΔvapH(H2O)=37.3kJmol−1

Calculate the entropy change at 373 K for the following transformation.
$H_{2} O (I, 1.01325 b a r) \to H_{2} O (g, 0.101325 b a r)$ . Given $Δ_{v a p} H (H_{2} O) = 37.3 k J {m o l}^{- 1}$