At 25 degree C, the equilibrium constant for the reaction below is 5.9 times 10^

Question

At 25 degree C, the equilibrium constant for the reaction below is 5.9 times 10^13. Suppose a container is filled with nitrogen dioxide at an initial partial pressure of 0.89 atm. Calculate the partial pressures of all three gases after equilibrium is reached at this temperature. (Oxtoby 14.34) At 1330 K, germanium(ll) oxide (GeO) and tungsten (VI) oxide (W_2O_6) are both gases. The equilibrium constants for reactions i and ii are 7 times 10^3 and 3.8 times 10^4. Using this information, calculate K for reaction iii. (Oxtoby 14.20)

Explanation / Answer

Q1.

Kp = 5.9*10^-13

Kp = P-N2^2 * P-O2 / (P-NO2)^2

initally

P-NO2 = 0.89 atm

P-O2 = 0

P-NO = 0

in equilibrium

P-NO2 = 0.89 atm -2x

P-O2 = 0 + x

P-NO = 0 + 2x

substitute

Kp = P-N2^2 * P-O2 / (P-NO2)^2

5.9*10^-13 =(2x)^2 *x / (0.89-2x)^2

(5.9*10^-13 ) = 4x^4 / (0.89^2)

x^3 = (5.9*10^-13 ) * (0.89^2) / 4

x = 1.168*10^-13)^(1/3) = 0.00004888184

P-NO2 = 0.89 -2*0.00004888184 = 0.8899

P-O2 = 0 + x = 0.00004888184

P-NO = 0 + 2x = 2*0.00004888184 = 0.00009776368

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