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For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is (Given: In 10 = 2.303 and log 2 = 0.3010)
- a)1.12
- b)2.43
- c)3.32
- d)33.31
Correct answer is option 'C'. Can you explain this answer?
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For a first order reaction, the time required for completion of 90% re...
First Order Reaction and Half Life
A first-order reaction is a chemical reaction that depends on the concentration of one reactant raised to the power of one, hence the name "first-order." A common example of a first-order reaction is the decay of radioactive isotopes. The rate of a first-order reaction is proportional to the concentration of the reactant. This means that as the concentration of the reactant decreases, the rate of the reaction also decreases.
The half-life of a reaction is the amount of time it takes for half of the reactant to be consumed. The half-life of a first-order reaction is defined by the equation:
t1/2 = (ln 2) / k
where t1/2 is the half-life, k is the rate constant, and ln is the natural logarithm.
90% Completion Time and Half Life
The question asks for the time required for completion of 90% reaction, which can be expressed as:
ln (Co/C) = k*t
where Co is the initial concentration, C is the concentration at time t, and k is the rate constant. Solving for t, we get:
t = (1/k) * ln (Co/C)
Now, we are told that this time is x times the half-life of the reaction. That is:
t = x * t1/2
Substituting the expression for t1/2 from above, we get:
x * (ln 2) / k = (1/k) * ln (Co/C)
Simplifying, we get:
ln (Co/C) = x * ln 2
Taking the antilog of both sides, we get:
Co/C = 2^x
Finally, solving for x, we get:
x = log2 (Co/C)
Using the given value of ln 10 and log 2, we get:
x = log10 (Co/C) / log10 2 = 3.32
Therefore, the value of x is 3.32.
A first-order reaction is a chemical reaction that depends on the concentration of one reactant raised to the power of one, hence the name "first-order." A common example of a first-order reaction is the decay of radioactive isotopes. The rate of a first-order reaction is proportional to the concentration of the reactant. This means that as the concentration of the reactant decreases, the rate of the reaction also decreases.
The half-life of a reaction is the amount of time it takes for half of the reactant to be consumed. The half-life of a first-order reaction is defined by the equation:
t1/2 = (ln 2) / k
where t1/2 is the half-life, k is the rate constant, and ln is the natural logarithm.
90% Completion Time and Half Life
The question asks for the time required for completion of 90% reaction, which can be expressed as:
ln (Co/C) = k*t
where Co is the initial concentration, C is the concentration at time t, and k is the rate constant. Solving for t, we get:
t = (1/k) * ln (Co/C)
Now, we are told that this time is x times the half-life of the reaction. That is:
t = x * t1/2
Substituting the expression for t1/2 from above, we get:
x * (ln 2) / k = (1/k) * ln (Co/C)
Simplifying, we get:
ln (Co/C) = x * ln 2
Taking the antilog of both sides, we get:
Co/C = 2^x
Finally, solving for x, we get:
x = log2 (Co/C)
Using the given value of ln 10 and log 2, we get:
x = log10 (Co/C) / log10 2 = 3.32
Therefore, the value of x is 3.32.
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For a first order reaction,
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For a first order reaction, the time required for completion of 90% reaction is x times the half life of the reaction. The value of x is (Given: In 10 = 2.303 and log 2 = 0.3010)a)1.12b)2.43c)3.32d)33.31Correct answer is option 'C'. Can you explain this answer?
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