The decomposition of solid ammonium carbamate, (NH_4)(NH_2 CO_2), to gaseous amm

Question

The decomposition of solid ammonium carbamate, (NH_4)(NH_2 CO_2), to gaseous ammonia and carbon dioxide is an endothermic reaction. (NH_4)(NH_2 CO_2)(s) 2NH_3 (g) + CO_2 (g) When solid (NH_4)(NH_2 CO_2) is introduced into an evacuated flask at 25 degree C, the total pressure of gas at equilibrium is 0.116 atm. What is the value of K_p at 25 degree C? Given that the decomposition reaction is at equilibrium, how would the following changes affect the total quantity of NH_3 in the flask once equilibrium is reestablished? Adding CO_2 Adding (NH_4)(NH_2 CO_2) Removing CO_2 Increasing the total volume Adding neon Increasing the temperature

Explanation / Answer

The reaction is

NH4NH2CO2-----> 2NH3 + CO2

The contribution to presure is from NH3 and CO2

Mole fractions

NH3= 2/3 and that of CO2= 1/3

Partial pressure = Mole fraction * total pressue =

NH3 partial pressue = (2/3)*0.116 =0.0773 atm and that of CO2= (1/3)*0.116= 0.0386

Equilibrium contant Kp = [partial pressure of NH3]2* partial pressure of CO2= (0.0773)2*0.0386 =0.000231

1. Adding CO2 will increase the number of moles of CO2 and hence backward reaction proceeds

2. Addition of NH4NH2CO2 will not have an effect on the equilibrium

3. Removing CO2 favors forward reaction

4. Increasing total volume proceeeds in a direction where quantity of Ammonia and CO2 increases. That is forward reaction is favored.

5. Addiin neon will add not have an impact since it is not partcipating in the reaction.

6. The reaction is exothermic, Hence increase in temperatur proceeds in the endothermic reaction. Hence, reverse reaction takes place.

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