Question

Hydrogen peroxide (H₂O₂ ) decomposes according to the equation

$2{\mathrm{H}}_2{\mathrm{O}}_2\rightleftharpoons 2{\mathrm{H}}_2\mathrm{O}(\mathrm{I})+{\mathrm{O}}_2(\mathrm{g})$

From the following data at $25^{{}^{\circ }}\text{C}$ calculate the value of K_p at 400 K for the above reaction, $\Delta {\text{H}}^{{}^{\circ }}=-196\cdot 0\text{ KJ }\Delta {\text{S}}^{{}^{\circ }}=125\cdot 65\text{J/K}$ .

[Given: $10^{0.15}=1.41$ ]

• A. 0.14 x 10³²
• B. 0.14 x 10^-32
• C. 0.14 x 10³
• D. 1.3 x 10¹⁵

Answer 1

Answer: (A)

$\Delta {\text{G}}^{{}^{\circ }}=+\Delta {\text{H}}^{{}^{\circ }}-\text{T}\Delta {\text{S}}^{{}^{\circ }}$

$=-196000-400\left(125\cdot 65\right)$

$=-196000-50260$

$=-246260\text{J}=-2\cdot 303\times 8\cdot 314\times 400\text{ log}{\text{ K}}_{\text{p }}$

$\text{log }{\text{K}}_{\text{p}}=\frac{-246260}{-7658}=32\cdot 15$

${\mathrm{K}}_{\mathrm{p}}=\mathrm{Antilog}(32\cdot 15)$

$=0.14\times 10^{32}$