For the process: H 2 O ( l ) → H 2 O ( g ) , Δ G 300 K = 9.00 kJ mol − 1 . If Δ H [ J m o l − 1 ] = 60000 + 5.0 × 1 0 − 3 ( T / K ) 2 , calculate Δ G 400 K (in kJ mol⁻¹).

Question

For the process: H 2 ​ O ( l ) → H 2 ​ O ( g ) , Δ G 300 K ​ = 9.00 kJ mol − 1 . If Δ H [ J m o l − 1 ] = 60000 + 5.0 × 1 0 − 3 ( T / K ) 2 , calculate Δ G 400 K ​ (in kJ mol⁻¹).

TestHub · Accepted Answer

The Gibbs-Helmholtz equation is given by:

$\frac{d}{d T} (\frac{Δ G}{T}) = - \frac{Δ H}{T ^{2}}$

Substitute the given $Δ H$ expression:

$\frac{d}{d T} (\frac{Δ G}{T}) = - (\frac{60000}{T ^{2}} + 5.0 \times 1 0^{- 3})$

Integrate both sides:

$\frac{Δ G}{T} = \frac{60000}{T} - 5.0 \times 1 0^{- 3} T + C$

Find $C$ using the conditions at $300 K$ ( $Δ G = 9000 J m o l^{- 1}$ ):

$\frac{9000}{300} = \frac{60000}{300} - (0.005 \times 300) + C$

$30 = 200 - 1.5 + C$

$C = 30 - 200 + 1.5$

$C = - 168.5$

Calculate at $400 K$ :

$\frac{Δ G _{400}}{400} = \frac{60000}{400} - (0.005 \times 400) - 168.5$

$\frac{Δ G _{400}}{400} = 150 - 2 - 168.5$

$\frac{Δ G _{400}}{400} = - 20.5$

The final answer is:

$Δ G_{400} = - 20.5 \times 400 = - 8200 J m o l^{- 1} = - 8.20kJmo l^{- 1}$

Chemistry - Thermodynamics - II Question with Solution | TestHub

Doubts & Discussion