When a graph between log K and 1/T is drawn a straight line is obtained. The temperature at which line cuts y-axis and x-axis.a)0, Ea/2.303 R log Ab)∞, Ea/(R ln A)c)0, log Ad)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

When a graph between log K and 1/T is drawn a straight line is obtained. The temperature at which line cuts y-axis and x-axis.a)0, Ea/2.303 R log Ab)&#8734;, Ea/(R ln A)c)0, log Ad)None of theseCorrect answer is option 'B'. Can you explain this answer?

Answer

When line cuts y axis

When it cut X -axis
log K = 0

Answer

Logk= logA-E/RT
it looks y=MX+ c
for cut y axis x=0 which can possible only for T=infinity
for cut x axis y=o T=E/R*lnA

Answer

To understand why the correct answer is option 'B', let's break down the given information and the concepts involved:

1. Graph between log K and 1/T:
- The given graph represents the relationship between the logarithm of the equilibrium constant (log K) and the reciprocal of temperature (1/T).
- This type of graph is commonly used to determine the activation energy (Ea) and the pre-exponential factor (A) in the Arrhenius equation.

2. Straight line:
- The given information states that when the graph is plotted, a straight line is obtained.
- This implies that the relationship between log K and 1/T is linear.

3. Cutting the y-axis:
- When a line cuts the y-axis, it means that the x-coordinate is zero.
- In this case, cutting the y-axis indicates the temperature at which log K is zero.

4. Cutting the x-axis:
- When a line cuts the x-axis, it means that the y-coordinate is zero.
- In this case, cutting the x-axis indicates the reciprocal of temperature (1/T) at which log K is zero.

Now, let's analyze the options given:

a) 0, Ea/2.303 R log A:
- This option does not provide any specific values for cutting the y-axis or x-axis. It mentions the activation energy (Ea) and the pre-exponential factor (A), but these values are not directly related to the given graph.

b) Ea/(R ln A):
- This option suggests that the temperature at which the line cuts the y-axis is zero, and the reciprocal of temperature at which the line cuts the x-axis is equal to the activation energy divided by the natural logarithm of the pre-exponential factor.
- As the given graph is linear, the line will intersect the y-axis at zero, and the x-axis at a specific value determined by this equation. Therefore, this option seems to be correct.

c) 0, log A:
- This option suggests that the temperature at which the line cuts the y-axis is zero, and the reciprocal of temperature at which the line cuts the x-axis is equal to the logarithm of the pre-exponential factor.
- However, the given graph does not provide any information about the value of the pre-exponential factor (A), so this option cannot be confirmed.

d) None of these:
- This option suggests that none of the given options are correct.
- However, based on the analysis above, option 'B' seems to be the most reasonable and consistent with the given information.

Therefore, the correct answer is option 'B' - Ea/(R ln A), as it provides a logical explanation for the temperature at which the line cuts the y-axis and x-axis in the given graph.

Similar Class 12 Doubts

Straight line graph for first order reaction is obtained between

The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then,

Can you explain the answer of this question below:The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then, A:Temperature T2is lessB:Temperature T2is moreC:T1is same as T2D:None of the aboveThe answer is c.

Select the correct statement(s).

Top Courses for Class 12

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams