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

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percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally