Above what temperature does the following reaction become nonspontaneous? 2 h2s(g) + 3 o2(g) → 2 so2(g) + 2 h2o(g) δh = -1036 kj; δs = -153.2 j/k

Above 6762 K the reaction becomes nonspontaneous.

Explanation:

Let's consider the following equation.

2 H₂S(g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(g)

This reaction will become nonspontaneous when the Gibbs free energy (ΔG) is higher than zero. ΔG is related to the enthalpy of the reaction (ΔH) and the entropy of the reaction (ΔS) through the following expression.

ΔG = ΔH - T.ΔS

If ΔG must be positive, then

ΔH - T.ΔS > 0

ΔH > T.ΔS

-1036 × 10³ J > T . (-153.2 J/K)

T > 6762 K

the answer is b. dilution

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.