(a) 2NO(g) + O₂(g) ⇄2NO₂(g)kp
(b) 2N₂O(g)⇄2NO(g) + N₂(g) kp
(c) N₂(g) + O₂(g)⇄ 2NO(g) kp
Now A is
2NO +O₂⇄2NO₂
ΔG° =ΔG° products - ΔG reactants
=2× 51.3-(256.6)
-70.6kJ/mol.
ΔG° = -RT Inkp
-70.6 = -8.314 ×10⁻³ ˣ 298.15 ˣInkJ
InkJ = 28.48
kp=2.34 ˣ 10¹²
B is
ΔG° = 2× 86.6 - 2 × 104.2 = -35.2
-35.2 = 8.314 × 10⁻³ ˣ 298.15 ˣInkJ
InkJ = 14.2
kp = 1.47ˣ 10⁶
C is
It is also similar
kp = 4.62 ˣ 10⁻³I
Answer:
and
Explanation:
Given:
- first charge,
- second charge,
- position of first charge,
- position of second charge,
Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.
<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>
- since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.
Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.
and
Answer:
B
Explanation:
The impulse experienced by an object is the force•time.
Answer: equation for the reaction is given below
PCL2+PCL3=PCL5
Where pcl2=0.40atm,pcl3=0.27atm
Pcl5=0.0029atm
Using ∆G=-RTin(PCL5/PCl2*PCL3)
Where R=8.314J/K/mol and T=298K
∆G=-8.314*298in(0.0029/0.40*.27)
∆G=8962.6J/mol
Explanation:
Answer:
1.503 J
Explanation:
Work done in stretching a spring = 1/2ke²
W = 1/2ke²........................... Equation 1
Where W = work done, k = spring constant, e = extension.
Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.
Substitute into equation 1
W = 1/2(26)(0.34²)
W = 13(0.1156)
W = 1.503 J.
Hence the work done to stretch it an additional 0.12 m = 1.503 J