The question is incomplete, the complete question is;
1. Given: 2A(g) <-> 2B(g) + C(g). At a particular temperature, K = 1.6x104.
At a higher temperature, K = 1.8x10-5. Placing the equilibrium mixture in an ice bath (thus lowering the temperature) will...
cause [A] to increase
cause [B] to increase
gave no effect
cannot be determined
Answer:
cause [B] to increase
Explanation:
We have to have it behind our minds that; for an exothermic reaction, when the temperature is increased, the value of the equilibrium constant decreases. On the other hand, for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant.
In this particular case, increasing the temperature decreased the equilibrium constant from K = 1.6x104 to K = 1.8x10-5 meaning that the equilibrium constant decreased with increase in temperature. This means that the reaction is exothermic and a decrease in temperature will favour the forward reaction hence more B is produced.
Answer:
soapy water is not good for plants irrigation that is what I learn
After 1911 most scientists accepted<span> the </span>theory<span> that the </span>nucleus<span> of an </span>atom<span> was </span>very dense<span> and </span>very small<span> and </span>has<span> a </span>positive charge<span>. </span>
Answer:
There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement. Because the lone pair isn't counted when you describe the shape, SO2 is described as bent or V-shaped.
Explanation:
There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement. Because the lone pair isn't counted when you describe the shape, SO2 is described as bent or V-shaped.
Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;
If 
&
E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol
