When solid NH4HS is placed in a closed flask at 28 celsius the solid disscoates

Question

When solid NH4HS is placed in a closed flask at 28 celsius the solid disscoates according to the equation:

16. When solid HAHs is placed in a closed flask at 28°C, the solid dissociates according to the equation NH3(g)+ The total pressure of the equilibrium mixture is 0.766 atm. Determine at this temperature. a) 0.147 Kp b) 0.766 c) 0.587 d) 0.383 e) 1.53 17. Consider the following exothermic reaction: N2(g) 3 H2(g) which of the following changes would not increase the amount of NH3 produced from given quantities of N2 and H2? a) increase in P b) increase in T c) decrease in V d) remove some NH3 and reestablish equilibrium e) none of the above 18. The hydrogenation of ethylene occurs by the following equilibrium: C2H4(g) 2 H2(g) 2 CH4( K 1.64 at 1000 K A vessel is charged with a 1.00 atm sample of C2H4(g) and a 1.00 atm sample of H20g) Estimate the pressure of methane gas after the system has reached equilibrium. a) 0.067 atm b) 0.13 atm c) 0.26 atm d) 0.52 atm e) 0.74 atm 19. For the reaction SO2 (g) Cl2(g) 5 SO2Cl2(g), K 55.5 at a certain temperature. If 1.00 mole of so2(g) and 1.00 mole of Cl2(g) are placed in a 10.0-L container and allowed to come to equilibrium, what is the equilibrium concentration of SO2Cl2 a) 0.87 b) 0.066 c) 0.13 d) 0.034 e) 0.74 20. K 0.25 for 2 NOBr(g) 2 NO(g) Br2(g) At same T and P, what is the K for NO 1/2 Br2 S NOBr? a) 2.0 b 4.0 d) 0.63 c) 0.50 e) 1.0

Explanation / Answer

Total pressrue is sum of partial pressures of gases

Hence partial pressure of NH3+ partial pressure of H2S= 0.766

Hence 2x= 0.766, where x= partial pressrue of NH3(PNH3) = partial pressure of H2S(PH2S) as per the reaction.

Since 2x=0.766, x= 0.766/2= 0.383

Hence Kp= equilibrium constant= [PH2S][PNH3]=0.383*0.383= 0.147 ( A is correct)

2.

The reaction N2(g)+3H2---à2NH3 is exothermic reaction. when P is increased since P is proportional to no of moles. P increase leads to increase in no of moles. So the reaction proceeds in a direction where there is a decrease in no of moles. So the reaction proceeds in forward direction.

Since the reactino is exothermic increase in T favors endothermic direction and hence towards reactant side ( correct option)

When V is decreased, the reaction proceeds where there is an increase in volume which favors the reactants.

When NH3 is removed, there is a decrease in mole of NH3, KC= [NH3]2/ [N2][H2]3,

so mole N2 and H2 will have to react to form NH3. ( correct option)

3.

Let x= drop in pressure of C2H4 to reach equilibrium

At equilibrium PC2H4= 1-x and PH2= 1-2x and PCH4= 2x, P indicates partial pressure

Hence Kp = (PCH4)2/ (PC2H4* PH2)= 1.64

Hence 4x2/((1-x)*(1-2x)2= 1.64, x2/(1-x)*(1-2x)2= 1.64/4=0.41

When solved using excel, x= 0.262, hence PCH4= 2*0.262=0.524 ( d is correct)

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