the standard free energy change for a chemical reaction is +13.3 kj/ mol. What is the equilibrium constant for the reaction at 125 degree C ? (R= 8.314 j/mol ) ?
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To the answer above me, you don't need a reaction to answer this because K, the equilibrium constant depends solely on temperature. Make sure to convert 13.3 kJ to joules and the 125 degrees C to Kelvin because the units of R are in joules and Kelvin
鈭咷= -RT饾搧nK, then rearrange eqution to 鈭咷 / -RT = 饾搧nK
1.33 x 10^4 Joules / (-8.314 J/Kmol x 398 Kelvin) = 饾搧nK
-4.019375202 = 饾搧nK
e ^-4.019375202 = K
K = 0.0180
OR
鈭咷= -2.303RTlogK, then rearrange equation to 鈭咷 / -2.303RT = logK
1.33 x 10^4 / (-2.303 x 8.314J/Kmol x 398 Kelvin) = logK
-1.745277986 = logK
10^-1.745277986 = K
K= 0.0180
use whichever equation you learned or feel comfortable with.
鈭咷= -RT饾搧nK, then rearrange eqution to 鈭咷 / -RT = 饾搧nK
1.33 x 10^4 Joules / (-8.314 J/Kmol x 398 Kelvin) = 饾搧nK
-4.019375202 = 饾搧nK
e ^-4.019375202 = K
K = 0.0180
OR
鈭咷= -2.303RTlogK, then rearrange equation to 鈭咷 / -2.303RT = logK
1.33 x 10^4 / (-2.303 x 8.314J/Kmol x 398 Kelvin) = logK
-1.745277986 = logK
10^-1.745277986 = K
K= 0.0180
use whichever equation you learned or feel comfortable with.
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Something is missing here - like the reaction itself? It would seem that you have not given us enough information to answer this question.
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No