Exams › NEET › Chemistry

A first order reaction has a rate constant of 2.303 × 10⁻³ s⁻¹. The time required for 40 g of this reactant to reduce to 10 g will be [Given that log10 2 = 0.3010]

1. 602 s
2. 230.3 s
3. 301 s
4. 2000 s

Correct answer: 602 s

Solution

For a first-order reaction, the integrated rate law is t = (2.303/k) log10 (R0/R). Substituting k = 2.303 × 10⁻³ s⁻¹, R0 = 40 g, and R = 10 g, we get t = (2.303 / 2.303 × 10⁻³) × log10 (40/10) = 1000 × log10 4 = 1000 × (2 × 0.3010) = 602 s.

Related NEET Chemistry questions

• Why is catalyst added while preparing oxygen from decomposition of \( K C l O_{3} ? \)
• In the Arrhenius equation \( k=A e^{E / R T} \) rate will be constant at:
• As the concentration of reactants increases:
• Assertion The rate of reaction can also increase w.r.t its product if one of the product act as catalyst. Reason A catalyst lowers the activation energy of reactions. Read the above assertion and reason and choose the correct option regarding ¡t.
• Assertion Chlorine reacts more rapidly with \( \boldsymbol{H}_{2}, \) in comparision to \( \boldsymbol{O}_{\mathbf{2}} \) Reason \( D-C l \) bond is stronger in comparison to \( H-C l \) bond
• In the Haber process for the industrial manufacturing of ammonia involving the reaction \( N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g) \) at 200atm pressure in the presence of a catalyst, a temperature of about \( 500^{\circ} \mathrm{C} \) is used. This is considered as optimum temperature for the process because:

⚔️ Practice NEET Chemistry free + battle 1v1 →