Sodium carbonate can be made by heating sodium bicarbonate: 2NaHCO_3(s) rightarr

Question

Sodium carbonate can be made by heating sodium bicarbonate: 2NaHCO_3(s) rightarrow Na_2CO_3(s) + CO_2(g) + H_2O(g) Given that Delta H degree = 128.9 kJ/mol and Delta G = 33.1 kj/mol at 25 degree C, above what minimum temperature will the reaction become spontaneous under standard state conditions? A) 0.4 K B) 3.9 K C) 321 K D) 401 K E) 525 K For the following reaction at equilibrium, what effect (shift) would heat do to the formation of products? 2NOBr(g) 2NO(g), Delta HP_rxn = 30 kJ/mol For the following reaction at equilibrium, what effect (shift) would an increase in pressure have on the formation of products? 2SO_2 (g) + O_2 (g) 2SO_3(g) Write the mass-expression, Q_c, for the following chemical reaction. Sn^2+(aq) + 1/2 O_2(g) + 3 H_2 (l) SnO_2(s) + 2H_3O^+ (aq)

Explanation / Answer

27. 2NOBr <-----> 2NO + Br2   dH=+30 Kj/mole

The above reaction is an endothermic reaction. Hence on addition of heat that is increasing the temperature of the system, By Le Chatelier's principle, the reaction will tend to shift towards right and more product will be formed.

If the heat is removed from the system, the temperature of the system will lower down.As a result of this the reaction will shift to left hand side and less product will be formed

28) 2SO2+02 ---> 2SO3

To understand the effect of pressure on the above reaction we will once again apply Le Chatelier's principle.

To understand this let us first claculate the number of moles on left hand side and right hand side of the reaction

Number of moles on Left hand side =2+1=3

Number of moles on Right hand side=2

Now if we increase the pressure of the system, the equilibrium will tend to move in a direction where the number of moles are less. In this case the number of moles on left hand side are more than the number of moles on right hand side. Hence on increasing the pressure, the reaction will shift to right hand side and more amount of SO3 will be formed.

If we decrease the pressure of the system, the equilibrium will tend to move in a direction where the number of moles are more. In this case the number of moles on left side are more than the number of moles on right hand side. Hence on decreasing the pressure of the reaction, the reaction will shift on left hand side and as a result of this , less amount of SO3 will be formed.

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