For a reaction Delta H = 9.08 kJ mol^{-1} a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The reaction is endothermic because $\Delta H > 0$ $(9.08 \ kJ \ mol^{-1})$.To determine spontaneity,we use the Gibbs free energy equation: $\Delt

Vedclass · Accepted Answer

Spontaneous and Endothermic

Vedclass · Answer

(C) The reaction is endothermic because $\Delta H > 0$ $(9.08 \ kJ \ mol^{-1})$.To determine spontaneity,we use the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.For the reaction to be spontaneous,$\Delta G$ must be negative $(\Delta G $\Delta G = 9080 \ J \ mol^{-1} - T(35.7 \ J \ K^{-1} \ mol^{-1})$.Setting $\Delta G  254.34 \ K$.Since the reaction can be spontaneous at temperatures above $254.34 \ K$ and is endothermic,option $(C)$ is the most appropriate description.

For a reaction $\Delta H = 9.08 \ kJ \ mol^{-1}$ and $\Delta S = 35.7 \ J \ K^{-1} \ mol^{-1}$. Which of the following statements is correct for the reaction?

Similar Questions

For water,the value of $\Delta H_{vap}$ is $40.73 \, kJ \, mol^{-1}$ and the value of $\Delta S_{vap}$ is $109 \, J \, K^{-1} \, mol^{-1}$. At what temperature will water be in equilibrium with its vapor?

$A$ chemical reaction cannot occur at all if

Gibbs free energy $G$,enthalpy $H$ and entropy $S$ are interrelated as in

What is the nature of the reaction at $298 \ K$,if the entropy change and enthalpy change for a chemical reaction are $7.4 \ cal \ K^{-1}$ and $-2.5 \times 10^3 \ cal$,respectively?

For the melting of $3 \ mol$ of ice at $0^o \text{C}$,$\Delta G = \dots \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module