Calculate P [NO]_eq if P [NOCl]_eq = 0.33 atm, P [Cl_2]_eq = 0.50 atm, and K_p =

Question

Calculate P [NO]_eq if P [NOCl]_eq = 0.33 atm, P [Cl_2]_eq = 0.50 atm, and K_p = 1.9 times 10^-2. 2 NOCl(g) = 2 NO(g) + Cl_2 (g) 0.0042 atm 0.087 atm 0.30 atm 0.064 atm 1.7 atm Give the direction of the reaction, if K = 1. The forward reaction is favored. The reverse reaction is favored. Neither direction is favored. If the temperature is raised, then the forward reaction is favored. If the temperature is raised, then the reverse reaction a favored. Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the system? CuS_(s) + O_2 (g) = Cu(s) + SO_2 (g) The reaction will shift to the right in the direction of products. The equilibrium constant will decrease. The equilibrium constant will increase. No effect will be observed. The reaction will shift to the left in the direction of reactants. The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. H_2 (g) + Br_2 (g) = 2 HBr (g) K_c = 38 times 10^4 4 HBr (g) = 2H_2 (g) + 2 Br_2 (g) K_c = ? 5.1 times 10^-3 1.6 times 10^3 1.9 times 10^4 2.6 times 10^-5 6.9 times 10^-10 Define dynamic equilibrium. the rate of the reverse reaction is faster than the rate of the forward reaction the rate of the forward reaction is equals the rate of the forward reaction no reactants react the rate of the forward reaction is faster than the rate of the reverse reaction no products are formed

Explanation / Answer

Question 5

First, state Kp

Kp = P-NO^2 * P-Cl2 /(P-NOCl)^2

substitute values

1.9*10^-2 = P-NO^2 *0.5 /(0.33^2)

P-NO^2 = (1.9*10^-2) / (0.5) * (0.33^2) = 0.004138

P-NO = sqrt(0.004138) = 0.06432 atm

best answer is D

Q6.

If K = 1

bothl products and reactants are favoured, since they are in equilibrium

so

choose C, neither direction

Q7.

if we increase pressure, nothing will happen

Since we have 1 mol in the reagent and 1 mol in the product

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