For the reaction: 2H2S(g) + 3O2(g)2H2O(g) + 2SO2(g) deltaH = -1.04×103 kJ and S° = -153 J/K The e...

Question

Question

For the reaction:

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

deltaH = -1.04×103 kJ and S° = -153 J/K

The equilibrium constant, K, would be greater than 1 at temperatures (above, below) _____ Kelvin.

Select above or below in the first box and enter the temperature in the second box. Assume that H° and S° are constant.

Answer 1

Answer #1

For the reaction:

2H2S(g) + 3O2(g) ----------> 2H2O(g) + 2SO2(g)

deltaH = -1.04 ×10^3 kJ

S° = -153 J/K

delta Ho = T delta So

T = delta Ho / delta So

= 1040 / 153

= - 6.797 K

temperature below 6.80 K

Answer 2

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