MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Gibbs Free Energy & Spontaneity

3 / 2 O 2 g⟶ ...

Question

$\frac{3}{2} O_{2} (g) ⟶ O_{3} (g)$ $K_{p}$ for this reaction is $2.47 \times 10^{- 29}$ . At 298 K, $Δ_{r} G^{o}$ for conversion of oxygen to ozone will be:

A

100 kJ mol

^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

150 kJ mol

^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

163 kJ mol

^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

2303 kJ mol

^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 163 kJ mol $^{- 1}$
For the reaction,

$\frac{3}{2} O_{2} (g) ⟶ O_{3} (g)$

The free energy is calculated as

$Δ_{r} G^{0} = - R T ln K_{p}$

Where T=298K

R=8.314

$K_{p} =$ equilibrium constant = $2.47 \times 10^{- 29}$

$Δ_{r} G^{0} = - 8.314 \times 298 \times ln (2.47 \times 10^{- 29}) ⟹ Δ_{r} G^{0} = 163228.93 J = 163.2 K J$

Suggest Corrections

thumbs-up

0

Similar questions

Δ G^{\circ}

for conversion of oxygen to ozygen to ozone

\frac{3}{2} O_{2_{(g)}} \to O_{3_{(g)}}

K_{p}

for this conversion

2.47 \times 10^{- 29}

2 F e_{2} O_{3} (s) \to 4 F e (s) + 3 O_{2} (g);

Δ_{r} G^{o} = + 1487.0

m o l^{- 1}

2 C O (g) + O_{2} (g) \to 2 C O_{2} (g);

Δ_{r} G^{o} = - 514.4

m o l^{- 1}

Free energy change,

Δ_{r} G^{o}

for the reaction,

2 F e_{2} O_{3} (s) + 6 C O (g) \to 4 F e (s) + 6 C O_{2} (g)

Q. Calculate free energy change for the conversion of oxygen to ozone at 298 K, if

K_{p}

\frac{3}{2} O_{2} (g) \to O_{3} (g)

2.47 \times 10^{- 29} (a t m)^{- \frac{1}{2}} .

log 2.47=0.3926

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $- 890.3 k J m o l^{- 1} - 393.5 k J m o l^{- 1}$ , and $- 285.8 k J m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) $- 74.8 k J m o l^{- 1}$
(ii) $- 52.27 k J m o l^{- 1}$
(iii) $+ 74.8 k J m o l^{- 1}$
(iv) $+ 52.26 k J m o l^{- 1}$

Q. For the reaction:

4 N H_{3} (g) + 3 O_{2} (g) \to 2 N_{2} (g) + 6 H_{2} O (l)

298 K

. Given that heat of formation of

N H_{3} (g

H_{2} O (l)

- 46 k J m o l^{- 1}

- 286 k J m o l^{- 1}

respectively. Calculate heat of reaction at

298 K

.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Gibbs Free Energy

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Gibbs Free Energy & Spontaneity

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon