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

Derivation of Kp and Kc

The standard ...

Question

The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below:
$C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)$
$△ H_{300 K}^{o} = - 41.16 k J m o l^{- 1}$
$△ S_{300 K}^{o} = - 4.24 \times 10^{- 2} k J m o l^{- 1}$
Calculate $K_{p}$ (at 300 K.)

4.93 \times 10^{4}

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

8.93 \times 10^{- 4}

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

8.93 \times 10^{4}

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

None of the above

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

Open in App

Solution

The correct option is C $8.93 \times 10^{4}$
At 300 K: $△ G^{o} = △ H^{o} - T △ S^{o}$
$= - 41.16 - 300 \times (- 4.24 \times 10^{- 2})$
$= - 28.44 k J$
we know
$△ G^{o} = - 2.303 R T l o g K_{P}$
$- 28.44 = - 2.303 \times 8.314 \times 10^{- 3} \times 300 l o g K_{P}$
$K_{P} = 8.93 \times 10^{4}$

Suggest Corrections

Similar questions

C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)

△ H_{300 K}^{o} = - 41.16 k J m o l^{- 1}

△ S_{300 K}^{o} = - 4.24 \times 10^{- 2} k J m o l^{- 1}

K_{p}

C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)

△ H_{300 K}^{o} = - 41.16 k J m o l^{- 1}

△ S_{300 K}^{o} = - 4.24 \times 10^{- 2} k J m o l^{- 1}

K_{p}

Δ H_{f}^{o}

C O_{2 (g)}

C O_{(g)}

H_{2} O_{(g)}

- 393.5

- 110.5

- 241.8

m o l^{- 1}

C O_{2 (g)} + H_{2 (g)} \to C O_{(g)} + H_{2} O_{(g)}

C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)

Δ H_{300 K}^{o} = - 41.16 k J {m o l}^{- 1}

Δ S_{300 K}^{o} = - 4.24 \times 10^{- 2} k J {m o l}^{- 1}

Δ H_{1200 K}^{o} = - 32.93 k J {m o l}^{- 1}

Δ S_{1200 K}^{o} = - 2.96 \times 10^{- 2} k J {m o l}^{- 1}

K_{P}

300 K

1200 K

P_{C O} = P_{{C O}_{2}} = P_{H_{2}} = P_{H_{2} O} = 1

^{- 1}

^{- 1}

^{- 1},

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program