A reaction will be nonspontaneous under all conditions when

a. delta G is negative
b. delta H is negative and DS is positive
c. delta H is positive and DS is negative
d. delta H is positive independent of the value and sign of DS
e. delta S is positive independent of the value and sign of DH

?

dG = dH - TdS.

Go through each answer and see if it fits; for example, #1 can't be right because when dG is - the reaction is spontaneous.
#2.
If dH is - and dS is +, then the term TdS is - so dG is always - when this is true and the rxn is spontaneous. So scratch 2.
Etc.

e is not it.

dG = dH - TdS.
If dS is + then the term -TdS will be negative but some dH values (even though they are + values) still can give a - dG value and the reaction will be spontaneous.

so d then

no. c

To determine which option is correct, we need to understand the conditions for a reaction to be nonspontaneous.

A nonspontaneous reaction is one that does not occur naturally or without external intervention. It needs to be forced or driven to occur.

The driving force for a reaction is Gibbs free energy (delta G), which combines enthalpy (delta H) and entropy (delta S) into a single parameter.

For a reaction to be spontaneous, delta G must be negative. Therefore, a nonspontaneous reaction will have a positive or zero delta G.

Analyzing each option:

a. delta G is negative
If delta G is negative, the reaction is spontaneous, not nonspontaneous. Therefore, this option is incorrect.

b. delta H is negative and DS is positive
For a reaction to be spontaneous, delta H must be negative (exothermic process, releasing heat) and delta S must be positive (increase in disorder). So, this option is incorrect as well.

c. delta H is positive and DS is negative
In this case, both delta H and delta S have unfavorable signs. This indicates that the reaction absorbs heat from the surroundings (endothermic process) and decreases disorder. Thus, this option is correct. A reaction will be nonspontaneous if delta H is positive and delta S is negative.

d. delta H is positive independent of the value and sign of DS
This option is incorrect because delta H alone cannot determine if a reaction is spontaneous or nonspontaneous. It needs to be considered along with delta S.

e. delta S is positive independent of the value and sign of delta H
Similarly, this option is incorrect because delta S alone cannot determine if a reaction is spontaneous or nonspontaneous. It needs to be considered along with delta H.

In conclusion, the correct option is c. A reaction will be nonspontaneous when delta H is positive and delta S is negative.