The following graph for a reaction is given. What is the enthalpy of reaction ( $Δ_{r} H^{\circ}$ )?
[Take R = 2 cal $K^{- 1} mol^{- 1}$ ; the line makes an angle θ with the horizontal dashed line such that tan θ = 2000 (the line falls as 1/T increases)]

Question

The following graph for a reaction is given. What is the enthalpy of reaction ( $Δ_{r} H^{\circ}$ )?

[Take R = 2 cal $K^{- 1} mol^{- 1}$ ; the line makes an angle θ with the horizontal dashed line such that tan θ = 2000 (the line falls as 1/T increases)]

log Kp versus 1/T graph

TestHub · Accepted Answer

log K p ​ = -ΔH/(2.303 R) × (1/T) + constant. The line falls, so the slope = -tan θ = -2000. -ΔH/(2.303 × 2) = -2000, so ΔH = +2000 × 4.606 = +9212 cal = +9.212 kcal. A falling line means K decreases as 1/T rises, i.e. K increases with T: endothermic. Answer: (B).

Chemistry - Chemical Equilibrium Question with Solution | TestHub

Options:

Doubts & Discussion