Let Delta P for H_2 be +x (since it is initially zero, it can only be Increasing

Question

Let Delta P for H_2 be +x (since it is initially zero, it can only be Increasing), then by stoichiometry Delta P for I_2 is also +x and Delta P for HI must be -2x. Add this information to the table, use it to obtain expressions for P_eq in terms of x, and then solve for x. K = (P_H_2) (P_I_2)/(P_HI)^2 = (x)(x)/(0.200 - 2x)^2 = x^2/(0.200 - 2x)^2 = 0.0169 Hence, x/0.200 - 2x = (0.0169)^1/2 = 0.13 So: x = 0.13(0.200 - 2x) = 0.026 - 0.26x or 0.026 = x + 0.26x = 1.26x Therefore, x = 0.026/1.26 = 0.021 Thus the equilibrium partial pressure of each gas is: P_H_2 = P_I_2 = x = 0.021 atm

Explanation / Answer

Given solution is correct.

Equilibrium partial pressureof all gases.

PH2 = x = 0.021 atm

PI2 = x= 0.021 atm

PHI = 0.200-(0.021atm*2)

= 0.158 atm

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