Standard enthalpy and standard entropy change for the oxidation of ${N H}_{3}$ at $298 K$ are $- 382.64 k J {m o l}^{- 1}$ and $- 145.6 k J {m o l}^{- 1}$ respectively. Standard Gibbs energy change for the same reaction at $298 K$ is:

+ 339.3 k J {m o l}^{- 1}

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

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

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

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

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

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

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

Open in App

Solution

The correct option is D $- 393.3 k J {m o l}^{- 1}$
Standard enthalpy and standard entropy change for the oxidation of ${N H}_{3}$ at $298 K$ are $- 382.64 k J {m o l}^{- 1}$ and $- 145.6 J {m o l}^{- 1} K^{- 1}$ respectively.
Standard Gibbs energy change for the same reaction at $298 K$ is $- 393.3 k J {m o l}^{- 1}$
$Δ G^{0} = Δ H^{0} - T Δ S^{0}$
$Δ G^{0} = - 382.64 k J {m o l}^{- 1} - (298 K \times (- 145.6 J {m o l}^{- 1} K^{- 1}) \times 10^{- 3} k J / J)$
$Δ G^{0} = - 393.3 k J {m o l}^{- 1}$

Suggest Corrections

Similar questions

Q. Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are382.64 kJ mol

^{- 1}

and 145.6 JK

^{- 1}

mol

^{- 1}

, respectively. Standard Gibbs energy change for the same reaction at 298 K is:

Standard enthalpy and standard entropy for the oxidation of $N H_{3}$ at $298$ K are $- 382.64$ $kJ {mol}^{- 1}$ and $- 145.6$ $J {mol}^{- 1} K^{- 1}$ respectively. Standard free energy change (in ${kJ mol}^{- 1}$ ) for the same reaction at $298$ K is: