A particular chemical reaction has a negative delta H and a negative delta S. Wh

Question

A particular chemical reaction has a negative delta H and a negative delta S. Which statement is correct?

(A) The reaction is spontaneous at all temperatures.

(B) The reaction is non spontaneous at all temperatures.

(C) The reaction becomes spontaneous as temperature increases.

(D) The reaction becomes spontaneous as temperature decreases.

Why?

When solid NH4N03 is dissolved in water at 25°C, the temperature of the solution decreases, What is
true about the signs of delta H and delta S for this process?
Why?

For the process 02(g) ~ 20(g), delta H = +498 kJ. What would be predicted for the sign of delta Srxn and the
conditions under which this reaction would be spontaneous?

(A) positive                   at low temperatures only
(B) positive at high temperatures only
(C) negative    at high temperatures only
(D) negative    at low temperatures only
Why?

For which reaction, carried out at standard conditions, would both the enthalpy and entropy changes
drive the reaction in the same direction?
(A) 2H2(g) + 02(g) ~ 2H2O(I)    delta H=-571.l kJ
(B) 2Na(s) + CI2(g) ~ 2NaCI(s)            delta H=-822.0kJ
(C) N2(gj + 202(g) ~ 2NOlg)      delta H=+67.7kJ
(D) 2NH)(g) ~ N2(g) + 3H2(g) (l)          deltaH = +92.4kJ
Why

Explanation / Answer

1.

delta G= deltaH - T* deltaS

negative delta S multiplied by T overpowers negative delta H and we get a positive delta G. But if T decreases a lot, then we might get delta G negative. So option D is correct.

2.

Temperature decrease, that means heat is absorbed, i.e delta H is positive. No. of ions increases in the solution, delta S is also positive.

3.

No. of molecules on product side(2) are greater than that of reactant side(1). so Srxn is positive.

As delta H is positive, if T is high enough, we can get negative delat G. So option C is correct.

4.

for determining the sign of entropy, calculate the no. of molecules on both reactant and product side. If no. is greater for product, delta S is positive, and vice versa.

So to derive the equation delta H and delta S should be of opposite sign(see the equation).

So the correct option is C)

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