X2O(. s .)leftharpoons2X(. s .)+(1/2)O2Δ (Go)(. X2 O .)=-1.72 kJ(mol)- 1 For the above reaction, what is the value of equilibrium pressure of O2 at 25oC in Nm- 2 ?

Question

X2O(. s .)leftharpoons2X(. s .)+(1/2)O2Δ (Go)(. X2 O .)=-1.72 kJ(mol)- 1 For the above reaction, what is the value of equilibrium pressure of O2 at 25oC in Nm- 2 ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(ÎG)o=-2.303 RTlog PO2 (at equil)(1/2) -1720J=-2.303× 8.314J× 298log PO2(1/2) log PO2(1/2)=0.3⇒ P(1/2)=2Nm- 2 ∴ P=4 Nm- 2

Q. $X_{2} O_{(s)} ⇌ 2 X_{(s)} + \frac{1}{2} O_{2} Δ (G^{o})_{(X_{2} O)} = - 1.72 k J (m o l)^{- 1}$
For the above reaction, what is the value of equilibrium pressure of $O_{2}$ at $2 5^{o} C$ in $N m^{- 2}$ ?

$(Δ G)^{o} = - 2.303 RTl o g P_{O_{2} (a t e q u i l)}^{\frac{1}{2}}$
$- 1720 J = - 2.303 \times 8.314 J \times 298 l o g P_{O_{2}}^{\frac{1}{2}}$
$l o g P_{O_{2}}^{\frac{1}{2}} = 0.3 \Rightarrow P^{\frac{1}{2}} = 2 N m^{- 2}$
$∴ P = 4 N m^{- 2}$

Q. X2​O(s)​⇌2X(s)​+21​O2​Δ(Go)(X2​O)​=−1.72kJ(mol)−1 For the above reaction, what is the value of equilibrium pressure of O2​ at 25oC in Nm−2 ?

(ΔG)o=−2.303RTlogPO2​(atequil)21​​ −1720J=−2.303×8.314J×298logPO2​21​​ logPO2​21​​=0.3⇒P21​=2Nm−2 ∴P=4Nm−2

Q. $X_{2} O_{(s)} ⇌ 2 X_{(s)} + \frac{1}{2} O_{2} Δ (G^{o})_{(X_{2} O)} = - 1.72 k J (m o l)^{- 1}$
For the above reaction, what is the value of equilibrium pressure of $O_{2}$ at $2 5^{o} C$ in $N m^{- 2}$ ?

$(Δ G)^{o} = - 2.303 RTl o g P_{O_{2} (a t e q u i l)}^{\frac{1}{2}}$
$- 1720 J = - 2.303 \times 8.314 J \times 298 l o g P_{O_{2}}^{\frac{1}{2}}$
$l o g P_{O_{2}}^{\frac{1}{2}} = 0.3 \Rightarrow P^{\frac{1}{2}} = 2 N m^{- 2}$
$∴ P = 4 N m^{- 2}$