A schematic plot of ln keq versus inverse of temperature for a reaction is shown in the figure. The reaction must be

Question

A schematic plot of ln keq versus 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.

Tardigrade Admin · Accepted Answer

ln (k2/k1)=(Δ H/R) [(1/T1)-(1/T2)] ln (6/2)=(Δ H/R) [1.5×10-3-2×10-3] Δ H comes to be negative. Hence the reaction is exothermic.

Q. A schematic plot of ln $k_{e q}$ versus inverse of temperature for a reaction is shown in the figure. The reaction must be

exothermic

endothermic

one with negligible enthalpy change

highly spontaneous at ordinary temperature.

$l n \frac{k _{2}}{k _{1}} = \frac{Δ H}{R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$
$l n \frac{6}{2} = \frac{Δ H}{R} [1.5 \times 1 0^{- 3} - 2 \times 1 0^{- 3}]$
$Δ H$ comes to be negative. Hence the reaction is exothermic.

Q. A schematic plot of ln keq​ versus inverse of temperature for a reaction is shown in the figure. The reaction must be