Hydrogen peroxide (H_2O_2) decomposes according to the equation Calculate K_p fo

Question

Hydrogen peroxide (H_2O_2) decomposes according to the equation Calculate K_p for this reaction at 25degree C. Is this a spontaneous reaction at standard state conditions and 25 degree C? At 1500degree C the equilibrium constant for the reaction Calculate Delta G degree for this reaction at 1500Degree for this reaction at 1500 Degree C. Is this a spontaneous reaction at standard state conditions and 1500 Degree C? Calculate K_p at 298 K for the reaction What is favored at equilibrium? (products or reactants) How did you determine this?

Explanation / Answer

1. dGo = dHo - TdSo

with,

dHo = -98.2 kJ/mol

dSo = 0.0701 kJ/k.mol

T = 25 + 273 = 298 K

we get,

dGo = -98.2 - 298 x 0.0701 = -119.09 kJ/mol

The reaction is spontaneous at 25 oC.

Now,

dGo = -RTlnKp

-119090 = -8.314 x 298 lnKp

So Kp = 7.50 x 10^20

2. Kp = 1.4 x 10^-7

dGo = -RTlnKp

        = -8.314 x (1500+273) ln(1.4 x 10^-7)

        = 232.63 kJ/mol

the reaction is non-spontaneous

3. dGof = dGof(products) - dGof(reactants)

              = (-370.4 + 86.7) - (-300.4 + 51.8)

              = -35.1 kJ/mol

dGof = -RTlnKp

-35100 = -8.314 x 298 lnKp

So, Kp = 1.42 x 10^6

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