In a chemical reaction, the chemical equilibrium is a state of system in which the concentration of reactants and products do not change with time.
At equilibrium, the reaction seems to be stopped due to the simultaneous rate change in the concentration of reactants and products.
Equilibrium attains a dynamic situation indicating that: the rate of forward reaction is equal to the rate of backward reaction and there is no net change in concentration.
Le Chatliers principle:
If a system at equilibrium is disturbed, the equilibrium moves in such a way to counteract the change.
Factors affecting chemical equilibrium are as follows:
oConcentration
oTemperature
oPressure ( or volume)
oFactors Affecting Equilibrium:
(a)
Consider the given equilibrium reaction:
Increase
is the reactant of the given reaction. So, when there is an increase in the concentration of reactant, it causes an increase in the yield of .
(b)
Consider the given equilibrium reaction:
Increase
is the product of the given reaction. So, when there is an increase in the concentration of product, it causes a decrease in the yield of .
(c)
Consider the given equilibrium reaction:
Decrease
is the reactant of the given reaction. So, when there is a decrease in the concentration of reactant, it causes a decrease in the yield of .
(d)
Consider the given equilibrium reaction:
Decrease the volume of the container in which the reaction occurs.
Change in the volume, effects the yield of the reaction.
Volume is inversely proportional to the pressure.
Decrease in the volume, increases the pressure of the container.
When there is an increase in the pressure of the container, equilibrium shifts to the side in which there are less number of moles. In the reactant side, only 9 moles are present, so the equilibrium shifts to the reactant side.
Therefore, when there is a decrease in the volume, then there is also a decrease in the yield of .
Ans: Part aIncrease the yield of
Part bDecrease the yield of
Part cDecrease the yield of
Part dDecrease the yield of
Consider the following equilibrium reaction: 4NH3(g) + 5O2(g) <---> 4NO(g) + 6H2O(g) Heat = -904.4 kJ...
The following question refers to the reaction: 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g) The reaction as written is exothermic. What will happen if heat is added to the system? (A) The equilibrium will shift to the left (B) The equilibrium will shift to the right (C) The NO concentration will remain constant, but there will be more water vapour (D) The equilibrium position will not change if heat is added (E) The system increase in temperature Explanation will be...
Consider the following reaction: 4NH3 + 5O2 4NO + 6H2O Determine the following rates at a time when the rate of consumption of O2 is 1.26e-03 M/s. a) rate of consumption of NH3 M/s b) rate of formation of NO M/s c) rate of formation of H2O M/s d) rate of reaction M/s
Determine delta sub r H in Kj/mol for this reaction. 4No+6H2O --> 4NH3 +5O2 using the equations and the enthalpy change of the reactions given. N2 + O2 --> 2NO delta sub r H= 180.1 NH3 --> 1/2 N2 +1.5H2. delta sub r H= 54.3 2H2O --> 2H2 + O2 delta sub r H= 486.7
Calculate the ^Ho in (kj) for the following reaction : 4NH3 (s) + 5O2 (g) ---> 4NO (g) + 6H2O (l) ^Ho: NH3 - 46.0 NO 90.0 H2O - 286 A. - 1,172 B. - 1,716 C. - 150 D. - 1,356
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Consider the following reaction: 4NH3(g) + 502(g) – 4NO(g) + 6H2O(g); AH = -906 kJ Now answer the following two questions. 1. Which condition will release more heat? A. Combining 4.0 g of NH3 with excess O2 or B. Combining 5.0 g of O2 with excess NH3 2. How much heat will be released from your answer in question 1? Condition will release more heat and the value of this heat energy is
Consider the reaction4NH3(g) +5O2(g) ⇌ 4NO(g)+6H2O(g) where ΔS° rxn= 180.5 J/KUsing standard thermodynamic data (in the Chemistry References), calculate the entropy change of the surroundings and the universe at 25°C.
Part A Ammonia reacts with oxygen according to the equation 4NH3(g)+5O2(g)→4NO(g)+6H2O(g),ΔHrxn=−906 kJ Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH3. Part B What mass of butane in grams is necessary to produce 1.5×103 kJ of heat? What mass of CO2 is produced? Assume the reaction to be as follows: C4H10(g)+132O2(g)→4CO2(g)+5H2O(g),ΔHrxn=−2658 kJ
. Given the following reaction at 25o C. 4NH3(g) + 5 O2(g) 4NO(g) + 6H2O(l) ΔHo = -1168 kJ ΔHo f(NH3) = -46.2 kJ/mol; ΔHo f(H2O) = -285.8 kJ/mol. What is the standard enthalpy of formation of NO gas at 25o C?
Consider the following reaction: 4NH3 + 502 --> 4NO + 6H2O. In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of 02. Only 0.490 g of NO is actually produced. What is the percent yield of NO?