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Read MoreThe rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
It is given that T1 = 298 K
∴T2 = (298 + 10) K
= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 10°.
Therefore, let us take the value of k1 = k and that of k2 = 2k
Also, R = 8.314 J K - 1 mol - 1
Now, substituting these values in the equation:

= 52897.78 J mol - 1
= 52.9 kJ mol - 1
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
It is given that T1 = 298 K
∴T2 = (298 + 10) K
= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 10°.
Therefore, let us take the value of k1 = k and that o...
Step-by-step explanation:
• It is given that T1 = 298 K
•
• ∴T2 = (298 + 10) K
•
• = 308 K
This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.
This question is based on core NCERT concepts explained in the chapter and should be revised thoroughly before exams.
Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.
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