Home
Class 11
CHEMISTRY
A schematic plot of log K(eq) vs inverse...

A schematic plot of log `K_(eq)` vs inverse of temperature for a reaction is shown in the figure. The reaction must be:

A

Exothermic

B

Endothermic

C

One with negligible enthalpy change

D

Highly spontaneous at ordinary temperature

Text Solution

Verified by Experts

`K_(c )=Ae^(-DeltaH^(ɵ)//RT)`
`log_(10) K_(eq)=log_(10)A-(DeltaH^(ɵ))/(2.303RT)`
`Y=C+MX`
Slope of the line will be positive, when `DeltaH^(ɵ)= -ve`, i.e. the reaction is xeothermic.
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Concept Applicationexercise 7.1|53 Videos

  • CHEMICAL EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Ex 7.2|40 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    CENGAGE CHEMISTRY|Exercise Archives Subjective|15 Videos

  • CLASSIFICATION AND NOMENCLATURE OF ORGANIC COMPOUNDS

    CENGAGE CHEMISTRY|Exercise Analytical and Descriptive Type|3 Videos

Similar Questions

Explore conceptually related problems

A schematic plot of ln K_(eq) versus inverse of temperature for a reaction is shown below The reaction must be

A schematic plot of ln K_(eq) versus inverse of temperature for a reaction is shown below: The reaction must be:

A schematic plot of In K_(eq) versus inverse o ftemperature for a reaction is shown below the reaction must be:

The plot of deviation (delta) vs angle of incidence (i) for a prism is as shown in the figure. Find the angle of the prism (A).

The graph relates. In K_(eq.) vs.(1)/(T) for a reaction. The reaction must be:

The stopping potential vs frequency plot for a substance is shown in the figure. The work function of the substance is

The plot of concentration of the reactant vs time for a reaction is a straight line with a negative slope. This reaction follows