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The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about :
(c) Rate at 50°C/ Rate at T1°C = (2)T/T1 = (2)50/10 = 25
or rt+10/rt =2
rt+50/rt =(2)5 = 32 times
Consider the reaction:
Cl2(aq) + H2S(aq) → S(s) +2H+(aq) +2Cl-(aq)
The rate equation for this reaction is rate = k[Cl2][H2S] Which of these mechanisms is/are consistent with this rate equation?
A. Cl2 + H2S → H+ + Cl- +Cl+ + HS- (slow)
cl+ + HS- → H+ +Cl- + S (fast)
B. H2S H+ + HS- (fast equilibrium)
Cl2 + HS- → 2Cl- + H+ + S (slow)
(a) For [A]r = K[Cl2][H2S]
For [B]r = K[Cl2][HS-]
3A C + D, For this reaction it is observed that when initial concentration of A is 10 mole/lit then t1/2 value of this reaction is 40 min and when initial concentration of A is 20 mole / lit, t1/2 value has been changed into 20 min. Which of the following is true?
In this question, t1/2 ∝ 1/a
It shows that the order of reaction is two.
In the following reaction : XA → YB
log[ -d(A)/dt ] = log[-d(B)/dt] +0.3
where -ve sign indicates rate of disappearance the reactant. Then X : Y is
The given equation log[-d(A)/dt] = log[-d(B)/dt] + 0.3 relates the rates of disappearance of reactants A and B. The difference of 0.3 in the logs implies that the rate of disappearance of B is 2 times the rate of disappearance of A. Hence, the stoichiometric coefficients X and Y are 2 and 1 respectively.
The activation energies of the forward and backward reactions in the case of a chemical reaction are 30.5 and 45.4 KJ/mol respectively. The reaction is
If Ef < Eb then reaction is exothermic.
The t0.5 for the first order reaction.
PCl5(g) PCl3(g) + Cl2(g) is 20 min. The time in which the conc. of PCl5 reduces to 25% of the initial conc. is close to
K = (2.303/t)log100/25
2.303x0.301/20 = 2.303/tx2x0.301
t = 40 min
Consider a reaction A + B + C. If the initial concentration of A was reduced from 4M to 2M in 1 hour and from 2M to 1M in 0.5 hours, the order of the reaction is
For zero order reaction
K1 = x/t = 2/1 = 2M/h
K2 = x/t = 1/0.5= 2M/h
because, K1 and K2 are same. Hence, reaction follows zero order.
A gaseous reaction A2(g) → B(g) + 1/2C(g) show increase pressure from 100 mm to 120 mm in 5 minutes. The rate of disappearance of (A2) will be
AZ(g) → B(g) + 1/2C(g)
At t = 0 100 0 0
At t = 5min (100 - x) x x/2
At t = 5min
100+x/2 = 120
x=40
-dA2/dt = 40/5 = 8mm min-1
In a catalytic reaction involving the formation of ammonia by Haber's process N2+3H2 → 2NH3, the rate of appearance of NH3 was measured as 2.5 x 10-4 mol L-1S-1 The rate of disappearance of H2 will be
N2+3H2 → 2NH3,
-1/3dH2/dt = -1/2dNH3/dt
-dH2/dt = 1/2x3x2.5x10-4
-dH2/dt = 3.75 x 10-4 mol L-1S-1
For a chemical reaction, A products, the rate of reaction doubles when the concentration of A is increased by 4 times. The order of reaction is
Rate ∝ [A]x
1/2 = 2 ∝ (4)x
x = 1/2
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