A.) Kp= 1.22 at 193 C. Calulate the value of the Kc for the reaction at 193C. Wh

Question

A.) Kp= 1.22 at 193 C. Calulate the value of the Kc for the reaction at 193C. Where the change in the number of moles is 2.I know you use the formula kp=Kc(RT) n

so I did (.0821 X 466)^-2/1.22. But it's not correct and I'm not sure where I'm going wrong, please help!

B.) At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.

H2(g)+I2 --> 2HI(g)     KC=53.3

At this temperature, 0.800 mol of H2 and 0.800 mol of I2 were placed in a 1.00-L container to react. What concentration of HI is present at equilibrium?

Explanation / Answer

A)kp=Kc(RT) n ,T=193+273=466 ,delta n=2-2=0

1.22=Kc( 0.0821*466)^0

1.22=Kc*1

Kc=1.22

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