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**Q1. **Which statement is correct?
Solution

Increase in pressure or concentration brings in an increase in collision frequency as well as increase in effective collision. Recall that energy of activation is not at all related with exothermic or endothermic nature

**Q2.**The half life for a reaction … of temperature
Solution

Half-life depends upon rate constant and rate constant (K) varies with temperature as K=A∙e^(-E_n/RT);K increase with temperature. Also t_(1/2)∝1/K

**Q3. **For a given reaction, pressure of catalyst reduces the energy of activation by 2 kcal at 27℃. The rate of reaction will be increased by:
Solution

K1=Ae^(-Ea/RT)
K2=Ae^(-[E_a-2]/RT)
K1/K2 =e^(2/RT)=e^(2/2×〖10〗^(-3)×300)=28

**Q4. **What is the formula to find value of t_(1/2) for a zero order reaction?
Solution

T_(1/2)=T_50,x=R/2
〖∴ T〗_50=R/k_0
So 〖 T〗_50∝R
T_50∝R/k_0
Therefore, the formula of t_(1/2)for a zero order reaction is [R]_0/2k

**Q5.**In a reaction, the threshold energy is equal to:
Solution

Activation energy is the needed by reactant molecules to gain threshold energy level.

**Q6. **Consider the reaction 2A+B→product
When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is
Solution

2A+B→product
[B] is doubled, half-life didn’t change
Half-life is independent of change in concentration of reactant i.e., first order
First order w.r.t. to B
+-+
When [A] is doubled, rate increased by two times
⇒ First order w.r.t.A
Hence, net order of reaction =1+1=2
Unit for the rate constant=conc.^((1-n)) t^(-1)
=(mol^(-1) )^(-1).s^(-1)
L.mol^(-1) s^(-1)

**Q7.**Which of these does not influence the rate of reaction?
Solution

Nature and concentration of the reactants and temperature of the reaction influence the rate of reaction. But molecularity does not affect the rate of reaction as it includes the number of atoms, ions or molecules that must collide with one another to result into a chemical reaction.

**Q8.**If a plot of log_10C versus t give a straight line for a given reaction, then the reaction is
Solution

A graph between the log concentration (log c) of reactant and time t for the first order reaction gives a straight line, whose slope is equal to -k/2.303
log_10〖C_A 〗=-kt/2.303+log_10〖(C_A )_0 〗
Hence, the order of the above reaction is one.

**Q9.**If the reaction rate at a given temperature becomes slower then
Solution

Slow reaction rate indicates higher free energy of activation

**Q1.**Which statement is correct?

Solution

Increase in pressure or concentration brings in an increase in collision frequency as well as increase in effective collision. Recall that energy of activation is not at all related with exothermic or endothermic nature

Increase in pressure or concentration brings in an increase in collision frequency as well as increase in effective collision. Recall that energy of activation is not at all related with exothermic or endothermic nature

**Q2.**The half life for a reaction … of temperature

Solution

Half-life depends upon rate constant and rate constant (K) varies with temperature as K=A∙e^(-E_n/RT);K increase with temperature. Also t_(1/2)∝1/K

Half-life depends upon rate constant and rate constant (K) varies with temperature as K=A∙e^(-E_n/RT);K increase with temperature. Also t_(1/2)∝1/K

**Q3.**For a given reaction, pressure of catalyst reduces the energy of activation by 2 kcal at 27℃. The rate of reaction will be increased by:

Solution

K1=Ae^(-Ea/RT) K2=Ae^(-[E_a-2]/RT) K1/K2 =e^(2/RT)=e^(2/2×〖10〗^(-3)×300)=28

K1=Ae^(-Ea/RT) K2=Ae^(-[E_a-2]/RT) K1/K2 =e^(2/RT)=e^(2/2×〖10〗^(-3)×300)=28

**Q4.**What is the formula to find value of t_(1/2) for a zero order reaction?

Solution

T_(1/2)=T_50,x=R/2 〖∴ T〗_50=R/k_0 So 〖 T〗_50∝R T_50∝R/k_0 Therefore, the formula of t_(1/2)for a zero order reaction is [R]_0/2k

T_(1/2)=T_50,x=R/2 〖∴ T〗_50=R/k_0 So 〖 T〗_50∝R T_50∝R/k_0 Therefore, the formula of t_(1/2)for a zero order reaction is [R]_0/2k

**Q5.**In a reaction, the threshold energy is equal to:

Solution

Activation energy is the needed by reactant molecules to gain threshold energy level.

Activation energy is the needed by reactant molecules to gain threshold energy level.

**Q6.**Consider the reaction 2A+B→product When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

Solution

2A+B→product [B] is doubled, half-life didn’t change Half-life is independent of change in concentration of reactant i.e., first order First order w.r.t. to B +-+ When [A] is doubled, rate increased by two times ⇒ First order w.r.t.A Hence, net order of reaction =1+1=2 Unit for the rate constant=conc.^((1-n)) t^(-1) =(mol^(-1) )^(-1).s^(-1) L.mol^(-1) s^(-1)

2A+B→product [B] is doubled, half-life didn’t change Half-life is independent of change in concentration of reactant i.e., first order First order w.r.t. to B +-+ When [A] is doubled, rate increased by two times ⇒ First order w.r.t.A Hence, net order of reaction =1+1=2 Unit for the rate constant=conc.^((1-n)) t^(-1) =(mol^(-1) )^(-1).s^(-1) L.mol^(-1) s^(-1)

**Q7.**Which of these does not influence the rate of reaction?

Solution

Nature and concentration of the reactants and temperature of the reaction influence the rate of reaction. But molecularity does not affect the rate of reaction as it includes the number of atoms, ions or molecules that must collide with one another to result into a chemical reaction.

Nature and concentration of the reactants and temperature of the reaction influence the rate of reaction. But molecularity does not affect the rate of reaction as it includes the number of atoms, ions or molecules that must collide with one another to result into a chemical reaction.

**Q8.**If a plot of log_10C versus t give a straight line for a given reaction, then the reaction is

Solution

A graph between the log concentration (log c) of reactant and time t for the first order reaction gives a straight line, whose slope is equal to -k/2.303 log_10〖C_A 〗=-kt/2.303+log_10〖(C_A )_0 〗 Hence, the order of the above reaction is one.

A graph between the log concentration (log c) of reactant and time t for the first order reaction gives a straight line, whose slope is equal to -k/2.303 log_10〖C_A 〗=-kt/2.303+log_10〖(C_A )_0 〗 Hence, the order of the above reaction is one.

**Q9.**If the reaction rate at a given temperature becomes slower then

Solution

Slow reaction rate indicates higher free energy of activation

Slow reaction rate indicates higher free energy of activation