A mixture of 2.0 mol of CO(g) and 2.0 mol of H_2 O(g) was allowed to come to equ

Question

A mixture of 2.0 mol of CO(g) and 2.0 mol of H_2 O(g) was allowed to come to equilibrium in a 10.0-L flask at a high temperature. If K = 4.0, what is the molar concentration of H_2 (g) in the equilibrium mixture? The equation for the reaction is this, CO(g) + H_2 O(g) CO_2 (g) + H_2 (g) (A) 0.67 M (B) 0.40 M (C) 0.20 M (D) 0.13 M Consider this reaction. 2SO_3 (g) 2SO_2 (g) + O_2 (g) What is the correct K_P expression for this reaction? (A) K_P = P^2_SO P_O/P^2_SO (B) K_P = P_SO_1 P_O_1/P_SO (C) K_P = (2 P_SO)^2 P_O/(2 P_SO)^2 (D) K_P = 2 P^2_SO_1 P_O_1/2 P^2_SO When a sample of NO_2 is placed in a container, this equilibrium is rapidly established. 2NO_2 (g) N_2 O (g) If this equilibrium mixture is a darker color at high temperatures or at low pressures, which statement about the reaction is true? (A) The reaction is exothermic and NO_2 is darker in color N_2 O_4. B) The reaction is exothermic and N_2 O_4 is darker in color than NO_2. (C) The reaction is endothermic and NO_2 is darker in N_2 O_4. (D) The reaction is endothermic and N_2 O_4 is darker in color than NO_2. Consider this reaction. AB_3 (g) A(g) + 3B(g) What is the equilibrium constant expression if the initial concentration of AB_3 is 0.1 M and equilibrium concentration of A is represented by x? Assume the initial concentration of A are both zero. (A) x - 3x/0.1 - x (B) x - x^3/(0.1 - x)^3 (C) x middot x^3/(0.1 - 3x)^3 (D) x middot (3x)^3/0.1 - x

Explanation / Answer

10) Kp is nothing but equilibrium constant of chemical reaction it is ratio of product of partial presure raised to power of product to the product of partial presure raised to power of reactants

I.e. answer is (C)

11). Given reaction is an example of exothermic reaction i.e. as reaction proceeds heat is genrated from the reaction and the temparature of reaction mixture decreases as heat is released

in given reaction NO2 is darker than N2O4, equilibrium is shifted towards N2O4 side as temparature decreases and temparature increases reaction equilibrium shifted towards NO2 side, identified by dark colur of raction mass

Hence answer is (A) Reaction is exothermic NO2 is darker in colour than N2O4

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