Answer:
(1) order = 2
(2) R = K [A]²
Explanation:
Given the reaction:
A--------->Product
The rate constant relation for the reaction is given as:
R(i) = K [A]............(*)
Where R(I) is rate constant at different concentration of A.
Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results
0.011 = K [0.15] ⁿ.........(1)
0.044 = K [0.30]ⁿ .......(2)
0.177 = K [0.60]ⁿ .........(3)
Dividing (2) by (1) and (3) by (1)
Gives:
0.044/0.011 = [0.3/0.15]ⁿ
4 = 2ⁿ; 2² = 2ⁿ; n = 2
Similarly
0.177/0.011 = [0.60/0.15]ⁿ
16.09 = 4ⁿ
16.09 = 16 (approximately)
4² = 4ⁿ ; n = 2
Hence the order of the reaction is 2.
The rate law is R = K [A]²
<span>density is an intensive property of the matter, it doesnt change with the change of mass for the same substance, at the same pressure and temperature</span>
First find the no. of moles of NaOH :
<span>30/1000 = 0.3 dm3 so no. of moles = 0.3*0.5 = 0.15 moles </span>
<span>as NaOH reacts with HNO3 in a ratio of one to one, there must have been 0.15 moles of HNO3 too </span>
<span>moles/volume = concentration </span>
<span>volume= 15/1000 = 0.15 dm3 </span>
<span>concentration = 1.15/0.15 = 1 mol.dm-3 </span>
<span>The quicker way would be to realize that you used twice as much NaOH so the HNO3 had to be twice as strong</span>
Its running pressure in torr would be :
5.10 x 10^8 pa x 1kPa/1000 Pa x 760 / 101.325 kPa
= 5.10 x 10 ^5 x 76 x 10^-1 x 1.01325 x 10 ^-2
= 392.7375 x 10^2
= 3.927275 x 10^4
ope this helps
