Observation no.
Time (in min)
Px (in mm Hg)
1
0
800
2
100
400
3
200
200
At constant temperature and volume, X decomposes according to first order kinetics as follows:
2X(g)→3Y(g)+2Z(g)
where Px is the partial pressure of X.
Which of the following statement(s) is/are true for the given reaction?
| Observation no. | Time (in min) | Px (in mm Hg) |
| 1 | 0 | 800 |
| 2 | 100 | 400 |
| 3 | 200 | 200 |
At constant temperature and volume, X decomposes according to first order kinetics as follows:
2X(g)→3Y(g)+2Z(g)
where Px is the partial pressure of X.
Which of the following statement(s) is/are true for the given reaction?
The correct option is D When Px=700 mm Hg, the total pressure of the system is 950 mm Hg
Px is the partial pressure of X
P0−P=Px at time 't'
where,
P0 is initial pressure at t=0
P is pressure of reactant which is already converted to product at time 't'.
Applying intergrated rate law of first order kinetics,
k=1t ln(P0P0−P) (In terms of pressure )
k=2.303tlogPoPx
P0=800 mm Hg
At time, t=100 min
Px=400 mm Hg
k1=2.303100 log(800400)
k1=2.303100 log2=6.932×10−3 min−1
Similary, at time, t=200 min
k2=2.303200 log(800200)
k2=2.303200 log4=6.932×10−3 min−1
Since k value is same in both, hence verified that it must be following the first order kinetics.
Rate constant, k=6.932×10−3 min−1
(C)
At 75% completion of reaction
k=2.303t log10025
6.932×10−3=2.303tlog10025
6.932×10−3=2.303t log4
6.932×10−3=2.303t×0.6
t=200 min
(D)
2X(g)→3Y(g)+2Z(g)
t=0: 800 0 0
t=t: (800−2P) 3P 2P
When the partial pressure of X is 700 mm Hg
then,
800−2P=700
2P=100
P=50 mm Hg
Total pressure =800−2P+3P+2P
=800+3P
=800+3×50
=950 mm Hg
Px is the partial pressure of X
P0−P=Px at time 't'
where,
P0 is initial pressure at t=0
P is pressure of reactant which is already converted to product at time 't'.
Applying intergrated rate law of first order kinetics,
k=1t ln(P0P0−P) (In terms of pressure )
k=2.303tlogPoPx
P0=800 mm Hg
At time, t=100 min
Px=400 mm Hg
k1=2.303100 log(800400)
k1=2.303100 log2=6.932×10−3 min−1
Similary, at time, t=200 min
k2=2.303200 log(800200)
k2=2.303200 log4=6.932×10−3 min−1
Since k value is same in both, hence verified that it must be following the first order kinetics.
Rate constant, k=6.932×10−3 min−1
(C)
At 75% completion of reaction
k=2.303t log10025
6.932×10−3=2.303tlog10025
6.932×10−3=2.303t log4
6.932×10−3=2.303t×0.6
t=200 min
(D)
2X(g)→3Y(g)+2Z(g)
t=0: 800 0 0
t=t: (800−2P) 3P 2P
When the partial pressure of X is 700 mm Hg
then,
800−2P=700
2P=100
P=50 mm Hg
Total pressure =800−2P+3P+2P
=800+3P
=800+3×50
=950 mm Hg
