Which statement is true?
a reaction in which the entropy of the system increases can be spontaneous only if it is exothermic.
a reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic.
a reaction in which the entropy of the system decreases can be spontaneous only if it is endothermic.

A

Explanation:

The entropy of a system has to deal with how energy is stored in a system. It tends to looks at the degree of disorderliness in the system.

The spontaneity of a reaction deals with the possibilities of the reactions occurring, while an exothermic reaction is one in which energy is released to the surroundings, the change in enthalpy here is negative.

The change in entropy or the change in enthalpy cannot dictate if a reaction is feasible I.e spontaneous or not. What determines the spontaneity of a chemical reaction is the change in free energy. For a reaction to be spontaneous, this value must be negative.

The change in free energy, the enthalpy and entropy are connected by the equation below:

ΔG = ΔH - TΔS

Now back to the question, for the first option, we were told that entropy increased and system is exothermic meaning negative enthalpy and positive entropy. In this case , change in free energy will be negative over all temperatures and hence, the reaction is spontaneous.

To the second option, we have negative entropy and negative enthalpy. Here change in free energy will only be negative at low temperature, which of course is not the best answer here

To the third option, we have negative entropy and positive enthalpy. This means change in free energy is only negative at high temperatures which of course is not the best answer.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol