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At 1177 K the partial pressures of an equilibrium mixture of H2O, H2 and O2 are...

At 1177 K the partial pressures of an equilibrium mixture of H2O, H2 and O2 are 0.0340, 0.0035 and 0.0013 atm, respectively. Given that the chemical equation for the equilibrium is, 2 H2O(g) <----> 2 H2(g) + O2(g) calculate Kp and Kc.

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Expert Solution

Solution:

Given data,

T= 1177k

P _H2O = 0.0340

P _H2= 0.0035

P _O2 = 0.0013

Reaction - 2H₂O 2H_2(g) + O_2(g)

To calculate Kp

From the above reaction Kp = products / reactants

Kp = [P _H2 ] ² * [P _O2 ] / [ P _H2O ]²

substitute the above value to the formula

Kp = (0.0035)² * (0.0013) / (0.0340)²

Kp = 1.377* 10^-5 atm

To find the Kc

2H₂O 2H_2(g) + O_2(g)

here the reaction is at equlibrium 4 moles of H₂0 give 2 moles of hydrogen and 2 moles of oxygen so n will be 0

We know the relation between Kp and kc hence the formula will be

Kp = Kc(RT)ⁿ so lets find out Kc from this relation

So Kc = Kp/ (RT)ⁿ

where Kp = 1.337* 10^-5

R = 8.314 J/mol

T = 1177 k

n = 0

substitute the values to above formula

Kc = Kp/ (RT)ⁿ

= 1.377*10^-5 / (8.314 * 1177)

Kc = 1.407*10^-9 mol/L

queen_honey_blossom answered 6 months ago

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