A reaction has a standard free-energy change of -15.50 kJ mol^-1 (-3.705 kcal mol^-1). how is the equilibrium constant calculated when the reaction is at 25 degrees celsius?

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Answer 1 · 2018-03-20T18:17:03

It is done like this:

You use the expression:

$sf(DeltaG^@=-RTlnK_(eq))$

$sf(T=25^@C=298K)$

$sf(R=8.31color(white)(x)"J/K/mol")$

$sf(lnK_(eq)=-(DeltaG^@)/(RT))$

$sf(lnK_(eq)=-(-15.50xx10^3)/(8.31xx298))$

$sf(lnK_(eq)=6.259)$

For which:

$sf(K_(eq)=522.8)$

(kcal is an obsolete unit)