Reaction between compounds (A) and (B) occurs as follows
A (g) + 2B (g) → 2C (g)
Following results were obtained
(1) Exp. No. | (2) Initial concentration of [A] mol L-1 | (3) Initial concentration of [B] mol L-1 | (4) Initial rate of formation of [C] mol L-1 s-1 |
1. | 0.30 | 0.30 | 0.10 |
2. | 0.30 | 0.60 | 0.40 |
3. | 0.60 | 0.30 | 0.20 |
The correct rate- law for the said reaction is (r and k rate and rate constant respectively)
r = k[A]2B
r = k[A][B]2
r = k[A][B]
r = k[A]2[B]2
In the reaction 2NO2 N2O4; the rate of disappearance of NO2 is equal to
2 [NO2]2
2k1[NO2]2 - 2k2[N2O4]
2k2[NO2]2 - k2[N2O4]
(2k1 - k2)[NO2]
The rate constant for the raection 2N2O5 → 4NO2 + O2 is 3.0 × 10-5 s-1. If rate is 2.40 × 10-5, then concentration of N2O5 (in mol/L) is
1.4
1.2
0.04
0.8
The concentrations of the reactant A in the reaction A → B at different times are given below
Concentration (M) | Time (minutes) |
0.069 | 0 |
0.052 | 17 |
0.035 | 34 |
0.018 | 51 |
The rate constant of the reaction according to the correct order of the reaction is
0.001 M/min
0.001 min-1
0.001 min/M
0.001 M-1 min-1
The half-life of two samples are 0.1 and 0.8 s. Their respective concentration are 400 and 50 respectively. The order of the reaction is
0
2
1
4
Acid hydrolysis of ester is first order reaction and rate constant is given by
k =
where, V0, Vt and V∞ are the volume of standard NaOH required to neutralise acid present at a given time, if ester is 50% neutralised then
V∞ = Vt
V∞ = (Vt - V0)
V∞ = 2Vt - V0
V∞ = 2Vt + V0
Consider the following reaction,
The reaction is of first order in each diagram, with an equilibrium constant of 104. For the conversion of chair form to boat form = 4.5 × 10-8 m at 298 K with pre-exponential factor of 1012 s-1. Apparent rate constant (=kA/ kB) at 298 K is
4.35 × 104 s-1
4.35 × 108 s-1
4.35 × 10-8 s-1
4.35 × 1012s-1
In the reversible reaction,
the rate of disapperance of NO2 is equal to
2k1[NO2]2 - 2k2[N2O4]
2k1[NO2]2 - k2[N2O4]
(2k1-k2)[NO2]
A chemical reaction was carried out at 300 K and 280 K. The rate constants were found to be k1 and k2 respectively. Then
k2= 4k1
k2=2k1
k2=0.25k1
k2=0.5k1
C.
k2=0.25k1
Rate constant becomes double for every 10° rise in temperature. Hence, for 20°C rise in temperature, rate constant will become four times. i.e.,
k1= 4k2
k2=0.25k1