CH_3OH(g) rightarrow CO(g) + 2 H_2(g) Delta H degree = + 91 kJ/mol_ rxn 25-26 Th

Question

CH_3OH(g) rightarrow CO(g) + 2 H_2(g) Delta H degree = + 91 kJ/mol_ rxn 25-26 The reaction represented above goes essentially to completion. The reaction takes place in a riding, insulated vessel that is initially at 600 K 25. What can be inferred about Delta S degree for the reaction at 600K? (A) It must be positive, since the reaction is thermodynamically unfavorable at 600 (B) It must be negative, since there are more moles of products than reactants. (C) It must be positive, since Delta G degree is negative and Delta H degree is positive. (D) It must be negative, since Delta G degree is positive and Delta H degree is positive.

Explanation / Answer

We know delta G = delta H - T * delta S

Here we know delta H = 91 kJ/mol that is positive which means reaction is endothermic.

Also we know that reaction goes to completion. We know from the rules of thermidynamics that if the reaction is moving forward means it is thermodynamically favorable. In such case delta G has to be negative.

Now in the above equation we are sure that delta G is negative and delta H is positive. In order to get the delta G negative delta S must be neagtive such that T*delta S value is greater than delta H.

Hence Ans: C

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---