Given: N_2(g) + O_2(g) 2NO(g) K_c = 0.16 What would be the equilibrium concentra

Question

Given: N_2(g) + O_2(g) 2NO(g) K_c = 0.16 What would be the equilibrium concentration of N_2(g) if 0.12 mole of NO(g) was placed in a 1.00-L contain at this temperature? What would be the new equilibrium concentration of N_2(g) if the volume were halved at this same temperature? 0.50 M; 1.0 M 0.070 M; 0.12 M 0.050 M; 0.12 M 0.10 M; 0.10 M 0.050 M; 0.10 M Consider the reversible reaction at equilibrium at 392 degree C. 2A(g) + B(g) C(g) The partial pressures are found to be: A: 6.70 atm, B: 10.1 atm, C; 3.60 atm. What is the concentration of B a equilibrium? 1.13 M 0.015 M 0.185 M 0.200 M 1.64 M Dinitrogen tetraoxide exists in equilibrium with nitrogen dioxide: N_2O_4(g) 2NO_2(g). At a certain temperature, K = 6.98 Times 10^4. Which of the following statements is true about this reactions? the reaction will reach equilibrium rapidly the concentration of NO_2 is equal to the concentration of N_2O_4at equilibrium the rate of dissociation of N_2O_4 is equal to the rate of formation of NO_2 at equilibrium the rate constant for the forward reaction is equal to the rate constant for the reverse reaction at equilibrium two of the above From the following data at 25 degree C, H_2(g) + Cl_2(g) rightarrow 2HCl(g) Delta H degree = -185 kJ 2H_2(g) + O_2(g) rightarrow 2H_2 O(g) Delta H degree = -483.7 kJ Calculate Delta H degree at 25 degree C for the reaction below 4HCl(g) + O_2(g) rightarrow 2Cl_2(g) + 2H_2O(g)

Explanation / Answer

18.

N2 + O2 <==> 2 NO

KC = 0.16 = [NO]^2/[N2][O2]

0.16 = (0.12-2X)^2/(X^2)

X = 0.05 M

N2 = 0.05 M

at new equilibrium

[N2] = 0.05*2 = 0.1 M

answer: e

20. e.two of the above.

21.

Dhrxn = 2*185 - 483.7 = -113.7 Kj

answer: A

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