Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are-382.64 kJ mol-1 and -145.6 JK-1 mol-1 respectively. Standard Gibbs energy change for the same reaction at 298 K is

Question

Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are-382.64 kJ mol-1 and -145.6 JK-1 mol-1 respectively. Standard Gibbs energy change for the same reaction at 298 K is (A) -523.2 kJmol-1 (B) -221.1 Mmol-1 (C) -339.3 kJmol-1 (D) -439.3 kJmol-1

Tardigrade Admin · Accepted Answer

NH3(g) + (3/4) O2(g) arrow (1/2) N2(g) + (3/2) H2O (l) T = 298 K Δ H - 382.64 kJ mol-1 Δ S = - 145.6 JK-1 mol-1 = - 0.1456 kJ K-1 mol-1 Δ G = Δ H - TΔ S = - 382.64 kJ mol-1 - (298 K) × (0.1456 K JK-1 mol-1 = - 382.64 kJ mol-1 + 43.3888 kJ mol-1 = -339.25 kJ mol-1

Q. Standard enthalpy and standard entropy changes for the oxidation of ammonia at $298 K$ are $- 382.64 k J m o l^{- 1} an d - 145.6 J K^{- 1} m o l^{- 1}$ respectively. Standard Gibbs energy change for the same reaction at $298 K$ is

$- 523.2 k J m o l^{- 1}$

$- 221.1 M m o l^{- 1}$

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

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

Q. Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298K are−382.64kJmol−1and−145.6JK−1mol−1 respectively. Standard Gibbs energy change for the same reaction at 298K is

−523.2kJmol−1

−221.1Mmol−1

−339.3kJmol−1

−439.3kJmol−1

NH3​(g)+43​O2​(g)→21​N2​(g)+23​H2​O(l) T=298K ΔH−382.64kJmol−1 ΔS=−145.6JK−1mol−1 = −0.1456kJK−1mol−1 ΔG=ΔH−TΔS = −382.64kJmol−1−(298K) ×(0.1456KJK−1mol−1 = −382.64kJmol−1+43.3888kJmol−1 = −339.25kJmol−1

Q. Standard enthalpy and standard entropy changes for the oxidation of ammonia at $298 K$ are $- 382.64 k J m o l^{- 1} an d - 145.6 J K^{- 1} m o l^{- 1}$ respectively. Standard Gibbs energy change for the same reaction at $298 K$ is

$- 523.2 k J m o l^{- 1}$

$- 221.1 M m o l^{- 1}$

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

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