Calculate the free energy change for the following reaction at 300 K. 2CuO(s) → Cu2O(s) + (1/2)O2(g) Given, Δ H=145.6 kJ mol-1 and Δ S=116 Jk-1mol-1

Question

Calculate the free energy change for the following reaction at 300 K. 2CuO(s) → Cu2O(s) + (1/2)O2(g) Given, Δ H=145.6 kJ mol-1 and Δ S=116 Jk-1mol-1  (A)  110.8 kJ mol-1  (B)  221.5 kJ mol-1  (C)  55.4 kJ mol-1  (D)  145.6 kJ mol-1

Tardigrade Admin · Accepted Answer

Δ G=Δ H-TΔ S =145.6× 103-300× 116 =145.6× 103-34.8× 103 =110.8× 103J mol-1 =110.8 kJ mol-1

Q. Calculate the free energy change for the following reaction at $300 K$ .
$2 C u O (s) \to C u_{2} O (s) + \frac{1}{2} O_{2} (g)$
Given, $Δ H = 145.6 k J m o l^{- 1}$ and $Δ S = 116 J k^{- 1} m o l^{- 1}$

$110.8 k J m o l^{- 1}$

$221.5 k J m o l^{- 1}$

$55.4 k J m o l^{- 1}$

$145.6 k J m o l^{- 1}$

$Δ G = Δ H - T Δ S$
$= 145.6 \times 1 0^{3} - 300 \times 116$
$= 145.6 \times 1 0^{3} - 34.8 \times 1 0^{3}$
$= 110.8 \times 1 0^{3} J m o l^{- 1}$
$= 110.8 k J m o l^{- 1}$

Q. Calculate the free energy change for the following reaction at 300K. 2CuO(s)→Cu2​O(s)+21​O2​(g) Given, ΔH=145.6kJmol−1 and ΔS=116Jk−1mol−1

110.8kJmol−1

221.5kJmol−1

55.4kJmol−1

145.6kJmol−1