Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
Answer:
See explanation
Explanation:
The reaction to be considered is shown below;
H2CO3<------->CO2 + H2O
We know that when a constraint such as a sudden change in concentration, pressure or temperature is imposed on a reaction system in equilibrium, the system has to adjust itself by shifting in a particular direction in order to cancel the constraint.
Now, if we remove CO2, the equilibrium position must shift to the right by the decomposition of more H2CO3 to establish equilibrium again.
Answer: <span>Fiberglass
Hope this helps!</span>
Using the ideal gas equation:
PV/T = constant
Therefore, the possible outcomes of increased pressure are:
Decreased volume if temperature is kept constant.
Increased temperature if volume is kept constant.
<h2>Greetings</h2>
Fr+ and P3-
This is because the Fr atom has a negative charge and P3 atom has a negative charge, these two atoms have an electrostatic force of attraction because they both need full outer shell of electrons to be a stable atom.
<h2>Hope this helps!</h2>
