A reaction has an equilibrium constant of 8.5 * 103 at 298 K.
At...

Question

A reaction has an equilibrium constant of 8.5 * 103 at 298 K. 
At 755 K, the equilibrium constant is 0.65. Find ΔHrxn ° for the 
reaction.

Accepted Answer

Hi everyone here we have a question asking what is the change in entropy of the reaction? For reaction with equilibrium constants of 0.98 at Kelvin and 47.9 at 400 Kelvin. So the natural log of our constant plus the change in in therapy for the reaction Over our gas constant Times one over T one equals delta S. For the reaction over our gas constant Equals the natural log of K two plus the change in entropy for our reaction over our gas constant Times one over T two. So the change in entropy for our reaction Equals the natural log of K two over K one over one over T one -1 over T two times are our K one Equals 0.98. R t one Equals Kelvin. R k two equals 0.9. RT two equals 400 Kelvin. So our change in entropy for our reaction equals the natural log of 47. Over 0. over one over 298 Kelvin -1 over 400 Kelvin, which equals 37, 788.49 Jules. And to change that to kill jules, we're going to move the decimal point to the left three places. So 0.788 kg joules. And we're gonna round that to 38 killed jules. And that is our final answer. Thank you for watching. Bye

A reaction has an equilibrium constant of 8.5 * 103 at 298 K. At 755 K, the equilibrium constant is 0.65. Find ΔHrxn ° for the reaction.

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