(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
With that information, you can determine the object's speed.
Just divide the distance covered by the time to cover the distance.
If you also know the direction the object moved, then you can
determine its velocity. If you don't, then you can't.
Total distance = 36500 m
The average velocity = 19.73 m/s
<h3>Further explanation</h3>
Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration


State 2 : constant speed

State 3 : deceleration


Total distance : state 1+ state 2+state 3

the average velocity = total distance : total time

Mohs hardness scale. hope this helps
As we know by work energy theorem
total work done = change in kinetic energy
so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

initial the box is at rest at position x = x1
so initial kinetic energy will be ZERO
at final position x = x2 final kinetic energy is given as

now work done is given as

so we can say

so above is the work done on the box to slide it from x1 to x2