Consider the following equilibrium: 2NO_(g) + Br_2(g) + energy 2NOBr(g) The equi
ID: 485325 • Letter: C
Question
Consider the following equilibrium: 2NO_(g) + Br_2(g) + energy 2NOBr(g) The equilibrium will Shift lo the left as a result of adding a catalyst. removing NOBr. increasing the volume. increasing the temperature. consider the following equilibrium N_2(g) + O_2 + energy 2NO(g) when the temperature is increased, the equilibrium shifts to the left and K_eq increases, left and k_eq decreases. right and k_eq increases. right and K_eq decreases. Consider the following equilibrium: 2CO(g) + O_2(g) 2CO_2(g)+ energy Some CO_2 is added to the equilibrium system at constant volume a new equilibrium is established. Compared to the original equilibrium, the rates of the reactions for the new equilibrium have An indication that an equilibrium system favours the products is a large K_eq, positive delta H. one step mechanism. low activation energy.Explanation / Answer
Q.8
According to Le Chatelier’s principle, when we disturb and equilibrium by any mean, it tries to regain and as a result of which, several quantities are changed including, pressure, temperature, concentration, mol, etc.
The given reaction is endothermic reaction as energy is required to be added. So, it is favorable when we increase the temperature.
Catalyst dost not make any change in shifting an equilibrium, but it just the speed up the reaction.
When we remove NOBr(g), it will go the right side.
When we increase the volume, it will favor the direction where number of moles are more.
Number of moles of left side = moles of NO + moles of Br = 2 + 1 = 3
Number of moles of right side = moles of NOBr(g) = 2
Left side has more mole than right, so it will go to the left side and hence the answer is
C. increase the volume
Q. 9
N2(g) + O2(g) + energy -- > 2NO(g)
This is endothermic reaction and when we increase the temperature it becomes favorable to form product more and goes to right side.
When know the relation between concentration change and keq as keq has product at the numerator and hence it will increase the value of keq.
So, answer would be
C. Right and Keq increases.
Q. 10
Reaction:
2 CO(g) + O2(g) ---- > 2 CO2(g) + energy
When we add CO2 to the given reaction then it would go the right side.
When we add more mole and its volume is constant this means that it would prefer the direction of right product side.
And so, the correct answer is
A. Increase Increase
Q. 11
Keq is the quantity needed to tell the spontaneity of the reaction. Its value determines the direction of the reaction.
So the answer is A large Keq
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