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NEET Chemistry MCQ
Quiz 6
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Q.1
The rate constant of first order reaction is 3×10⁻⁶ per second. the initial concentration is 0.10M. The initial rate is ....[ AIIMS 1996]
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a) 3 ×10⁻⁷ Ms-1
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b) 3 ×10⁻⁸ Ms-1
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c) 3 ×10⁻⁵ Ms-1
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d) 6 ×10⁻⁹ Ms-1
Explanation
rate = k × (initial concentration) Answer : (a)
Q.2
The rate constant for the reaction N2O5(g) → 2NO2(g) + ½ O2(g) is 2.3×10⁻² sec-Which equation given below describes the change of [N2O5] with time [N2O5]0 and [N2O5]t? ... [ AIIMS 2004 ]
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a) [N2O5]0 = [N2O5]tekt
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b) [N2O5]t = [N2O5]0 + kt
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c) log10[N2O5]t = log10[N2O5]0 - kt
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d)
Explanation
Answer : (d)
Q.3
The rate constant for the hydrolysis reaction of an ester by dilute acid is 0.6931×10⁻³ s-The time required to change the concentration of eter from 0.04M to 0.01M is
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a) 6931 sec
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b) 1000 sec
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c) 4000 sec
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d) 2000 sec
Explanation
Answer : (d)
Q.4
The decay constant of 6C14 is 2.31 ×10⁻⁴ year-Its half-life is ...[ BHU 2000]
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a) 2×103 yrs
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b) 2.5×103 yrs
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c) 3×103 yrs
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d) 3.5×103 yrs
Explanation
Decay of nucleus is first order reaction use formula Answer : (c)
Q.5
For the reaction N2O5(g) → 2NO2(g) + ½O2(g), the value of rate of disappearance of N2O5 is given as 6.25×10⁻³ mol L-1s-The rate of formation of NO2 and O2 is given retrospectively as ...[ CBSE 2010]
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a) 6.25×10⁻³ mol L-1s-1 and 6.25×10⁻³ mol L-1s-1
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b) 1.25×10⁻² mol L-1s-1 and 3.125×10⁻³ mol L-1s-1
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c) 6.25×10⁻³ mol L-1s-1 and 3.125×10⁻³ mol L-1s-1
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d) 1.25×10⁻² mol L-1s-1 and 6.25×10⁻³ mol L-1s-1
Explanation
Answer : (b)
Q.6
A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as, k1 = 1.26×10⁻⁴ sec-1, k2 = 3.8 ×10⁻⁵ sec⁻¹ The percentage distribution of B and C are ...[ Kerala PMT 2004]
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a) 75% B and 25% C
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b) 80% B and 20% C
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c) 60% B nad 40% C
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d) 90% and 10% C
Explanation
Given reaction is parallel reaction ∴ %[C]= 100 - 76.83 = 23.17% Answer : (a)
Q.7
Rate of reaction for following reaction is 3×10⁻³ mol litre-1 min-What is the value of -d[B]/dt A(g) +3B(g) → 2C(g)
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c) 10-3
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a) 3×10⁻³
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b) 9×10⁻³
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d) 1.5×10⁻³
Explanation
Answer : (b)
Q.8
In the reaction : A + 2B → C + 2D the initial rate at t=0 was found to be 2.6×10⁻² M sec-What is the value of rate of disappearance of B at t=0 in M sec-1
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a) 2.6×10⁻² M sec-1
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b)5.2×10⁻² M sec-1
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c) 1.0×10⁻¹ M sec-1
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d) 6.5×10⁻³ M sec-1
Explanation
Answer : (b)
Q.9
The reaction, X → Y (product) follows first order kinetics. In 40 minutes, the concentration of X changes from 0.1M to 0.025M, then the rate of reaction when concentration of X is 0.01M is
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a) 3.47 × 10⁻⁴ M/min
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b) 3.47 × 10⁻⁵ M/min
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c) 1.73 ×10⁻⁴ M/min
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d) 1.73 ×10⁻⁵ M/min
Explanation
Use formula to find k Substitute R= 0.1M ,R0= 0.025M on simplification we get k=0.03466 Now rate = k × (concentration 0.01M) rate = 3.47 × 10⁻⁴ M/min Answer : (a)
Q.10
A drop of solution (volume 0.05 mL) contains 3×10⁻⁶ mole of H+. if the rate constant of disappearance of H+ is 107 mol litre- sec-1, how long would it take for H+ in the drop to disappear?
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a) 6×10⁻⁸ sec
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b) 6×10⁻⁹ sec
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c) 6×10⁻¹⁰ sec
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d) 6×10⁻¹² sec
Explanation
concentration of [H]+ = Number of moles /volume From the unit of rate constant, reaction is of zero order Now given 107 mole disappearance in 1 sec. ∴ 6×10⁻² mole disappearance in 6×10⁻⁹ sec Answer : (b)
Q.11
Rate constant of a reaction is 125 litre2 mol-2 sec- What is the order of reaction?
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a) First
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b) Second
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c) Third
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d) Zero
Explanation
Given unit of k =litre-2 mol-2 sec⁻¹ Comparing the unit 0f litre . n-1=2 n=3 Answer : (c)
Q.12
For the reaction 2N2O5 → 4NO2 + O2 rate and rate constant are 1.02×10⁻⁴ Ms-1 and 3.4×10⁻⁵ s-1 respectively then concentration of N2O5 at that time will be ...[ CBSE 2001]
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a) 17.732 M
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b) 3 M
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d) 3.4 ×105 M
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c) 1.02 ×10⁻⁴ M
Explanation
Rate = k[N2O5] 1.02×10⁻⁴ = 3.4 ×10-[N2O5] 1.02×10⁻⁴ = 3.4×10⁻⁵ [N2O5] [N2O5]=3M Answer : (b)
Q.13
For the decomposition of HI at 1000K 2HI → H2 + I2, the following data were obtained The oder of the reaction is
[HI],M
Rate of decomposition of HI(mol L
-1
s
-1
)
0.1
2.75×10
-8
0.2
11×10
-8
0.3
24.75×10
-8
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a) 0
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b) 1
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c) 1.5
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d) 2
Explanation
Only one reactant thus Rate = k[HI]n 11×10⁻⁸ = k[0.2]n ...(i) 2.75×10⁻⁸ = k[0.1]n ...(ii) from (i) and (ii) 4 =2n, n=2 Answer : (d)
0 h : 0 m : 1 s
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