Which one of the following is spontaneous at all temperatures?

Question

Which one of the following is spontaneous at all temperatures? (A)  H2(g) xrightarrow2H textatom Δ Ho=436 kJ, mol-1 Δ So=90.7 mol/K  (B)  (1/2)N2(g)+(1/2)O2(g) xrightarrowNO(g); Δ Ho=90.3 kJ,Δ So=3.0 kJ/mol/K  (C)  2NO2(g) xrightarrowN2O4(g) Δ Ho=-56.0 kJ , Δ So=-17.7 texteu  (D)  H2O2(g) xrightarrowH2O(l)+(1/2)O2(g) Δ Ho=-98.3 kJ, Δ So=80.0 kJ/mol/K

Tardigrade Admin · Accepted Answer

For a spontaneous process, Δ G must be negative. ⇒ Δ G=Δ H-TΔ S ∴ For negative value of Δ G at all temperatures, Δ H=-ve Δ S=+ ve

Q. Which one of the following is spontaneous at all temperatures?

$H_{2}_{(g)} 2 H_{atom}$ $Δ H^{o} = 436 k J, m o l^{- 1} Δ S^{o} = 90.7 m o l / K$

$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) NO (g);$ $Δ H^{o} = 90.3 k J, Δ S^{o} = 3.0 k J / m o l / K$

$2 N O_{2} (g) N_{2} O_{4} (g)$ $Δ H^{o} = - 56.0 k J, Δ S^{o} = - 17.7 eu$

$H_{2} O_{2}_{(g)} H_{2} O_{(l)} + \frac{1}{2} O_{2}_{(g)}$ $Δ H^{o} = - 98.3 k J, Δ S^{o} = 80.0 k J / m o l / K$

For a spontaneous process, $Δ G$ must be negative.
$\Rightarrow$ $Δ G = Δ H - T Δ S$
$∴$ For negative value of $Δ G$ at all temperatures,
$Δ H = - v e$ $Δ S = + v e$

Q. Which one of the following is spontaneous at all temperatures?

H2​(g)​​2Hatom​ ΔHo=436kJ,mol−1ΔSo=90.7mol/K

21​N2​(g)+21​O2​(g)​NO(g); ΔHo=90.3kJ,ΔSo=3.0kJ/mol/K

2NO2​(g)​N2​O4​(g) ΔHo=−56.0kJ,ΔSo=−17.7eu

H2​O2​(g)​​H2​O(l)​+21​O2​(g)​ ΔHo=−98.3kJ,ΔSo=80.0kJ/mol/K

For a spontaneous process, ΔG must be negative. ⇒ ΔG=ΔH−TΔS ∴ For negative value of ΔG at all temperatures, ΔH=−ve ΔS=+ve

$H_{2}_{(g)} 2 H_{atom}$ $Δ H^{o} = 436 k J, m o l^{- 1} Δ S^{o} = 90.7 m o l / K$

$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) NO (g);$ $Δ H^{o} = 90.3 k J, Δ S^{o} = 3.0 k J / m o l / K$

$2 N O_{2} (g) N_{2} O_{4} (g)$ $Δ H^{o} = - 56.0 k J, Δ S^{o} = - 17.7 eu$

$H_{2} O_{2}_{(g)} H_{2} O_{(l)} + \frac{1}{2} O_{2}_{(g)}$ $Δ H^{o} = - 98.3 k J, Δ S^{o} = 80.0 k J / m o l / K$

For a spontaneous process, $Δ G$ must be negative.
$\Rightarrow$ $Δ G = Δ H - T Δ S$
$∴$ For negative value of $Δ G$ at all temperatures,
$Δ H = - v e$ $Δ S = + v e$