Nitrogen and oxygen gases may react to form nitrogen monoxide. At 1500 degree C,

Question

Nitrogen and oxygen gases may react to form nitrogen monoxide. At 1500 degree C, K_c equals 1.0 times 10^-5 N_2 (g) + O_2 (g) 2 NO(g) 1030 mol N_2 and 0.030 mol O_2 are sealed in a 1.0 L flask at 1500 degree C, what is the concentration of NO(g) when equilibrium is established? a. 3.0 times 10^-7 M b. 4.7 times 10^-5 M c. 9.5 times 10^-5 M d. 3.0 times^10-2 M e. 9.1 times 10^1 M Sulfuryl chloride decomposes to sulfur dioxide and chlorine. SO_2 CI_2 (g) SO_2 (g) + CI_2 (g) K_c is 0.045 at 648 K. If an initial concentration of 0.075 M SO_2 CI_2 is allowed to equilibrate, what is the equilibrium concentration of CI_2? a. 0.0034 M b. 0.030 M c. 0.040 M d. 0.058 M e. 0.075 M Write a balanced chemical equation which corresponds to the following equilibrium constant expression. K = [NO_2^-][H_3 O^+]/[HNO_2] a. HNO_2 (aq) + H_2 O (l) NO_2^- (aq) + H_3 O^+ (aq) b. NO_2^- (aq) + H_3 O^+ (aq) HNO_2 (aq) + H_2 O(l) c. NO_2^- (aq) + H3 O^= (aq) HNO_2 (aq) d. H^+ (aq) + OH^- (aq) H_2 O(l) e. HNO_2 (aq) NO_2^- (aq) + H_3 O^- (aq) If the reaction quotient, Q, is equal to AT in a gas phase reaction, then a. the chemical system has reached equilibrium. b. the temperature must be increased for the reaction to proceed in the forward direction. c. the reaction will proceed in the forward direction until equilibrium is established. d. the reaction will proceed in the backward direction until equilibrium is established. e. the reaction will proceed in the direction that increases the number of gas phase particles.

Explanation / Answer

Q23.

Kc = 10^-5

N2+O2 = 2NO

We need concentrations, recall that M = mol/V , since V = 1 L, then assume mol = M

[N2] = mol/V = 0.03/1 = 0.03 M

[O2] = mol/V = 0.03/1 = 0.03 M

[NO] = 0

in equilibrium, apply extent of reactin "x"

[N2] = 0.03 -x

[O2] = 0.03 - x

[NO] = 0 + 2x

substitute in Kc expression:

Kc = [NO]^2/([N2][O2])

10^-5 = (2x)^2 / (0.03 -x)(0.03 -x)

10^-5 = (2x)^2 / (0.03 -x)^2

solve for x

sqrt(10^-5) = sqrt( (2x)^2 / (0.03 -x)^2)

0.00316227 = 2x / (0.03 -x))

0.00316227*0.03 - 0.00316227x = 2x

(2+0.00316227)x = 0.00316227*0.03

x = ( 0.00316227*0.03) / ((2+0.00316227)) = 0.00004735916

substitute in NO:

[NO] = 0 + 2x = 2*0.00004735916 = 9.47183*10^-5

nearest answer will be C, 9.5*10^-5 M

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