Question

A schematic plot of ln K_eq versus inverse of temperature for a reaction is shown below

The reaction must be:

Options

• highly spontaneous at ordinary temperature

• one with negligible enthalpy change 

• Endothermic

• Exothermic

Answer 1

Exothermic

Explanation:

`"K"_"f" = "A"_"f" e^(-"E"_"f"//"RT") and "K"_"b" = "A"_"b" e^(- "E"_"b"//"RT")`

We know that `"K"_"eq" = "K"_"f"/"K"_"b" => "K"_"eq" = "A"_"f"/"A"_"b" e^(- ("E"_"f" - "E"_"b")//"RT")`

`=> ln "K"_"eq" = ln  "A"_"f"/"A"_"b" - ("E"_"f" - "E"_"b")/"RT"`

`=> ln "K"_"eq" = (- Delta "H")/"R" (1/"T") + ln  "A"_"f"/"A"_"b"`

Comparing this equation with y = mx + c

m = `(- Delta "H")/"R" = + "ve"`

Thus, ΔH < 0 or reaction is exothermic.