Based on the thermodynamic functions of enthalpy and entropy, can an unfavorable reaction that has a positive ΔG at RT be made favorable by increasing the reaction temperature?Based on the thermodynamic functions of enthalpy and entropy, can an unfavorable reaction that has a positive ΔG at RT be made favorable by increasing the reaction temperature?

A. yes, but only if ΔH is positive and ΔS is negative
B. Yes, regardless of the sign of ΔH, if ΔS is positive, a reaction can be rendered favorable by increasing the temperature.
C. only if both ΔH and ΔS are negative
D. No, an unfavorable reaction cannot be rendered favorable.

B.Yes, regardless of the sign of ΔH, if ΔS is positive, a reaction can be rendered favorable by increasing the temperature

Explanation:

Given

∆G= ∆H- T∆S

If ∆G is to remain negative, then irrespective of the sign of ∆H, ∆S must be positive. If ∆S is positive, then increase in T will only lead a more negative value of ∆G. The positive ∆S maintains T∆S at a positive value that can be subtracted from the value of ∆H which is constant.

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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