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Viscosity

Mitty1990

Hi, I'm struggling to rearrange the Huggins and Kraemer equations to find both k' and k''.

Huggins: n red = n sp /c = [n] + k’ [n]2 c

Kraemer: n inh = ln n rel /c = [n ] + k” [n]2 c

Thank you :smile:

Reply 1

Plato's Trousers

Original post by Mitty1990

Huggins:

\eta_{red}= [\eta] + k' [\eta]^2c

so:

k'=\dfrac{\eta_{red} - [\eta]}{[\eta]^2c}

Kraemer:

\eta_{inh}= [\eta] + k'' [\eta]^2c

k''=\dfrac{\eta_{inh} - [\eta]}{[\eta]^2c}

Reply 2

Mitty1990

Original post by Plato's Trousers

Huggins:

\eta_{red}= [\eta] + k' [\eta]^2c

so:

k'=\dfrac{\eta_{red} - [\eta]}{[\eta]^2c}

Kraemer:

\eta_{inh}= [\eta] + k'' [\eta]^2c

k''=\dfrac{\eta_{inh} - [\eta]}{[\eta]^2c}

Thanks, but does the ln need to be in the Kraemer equation?

Reply 3

Plato's Trousers

Original post by Mitty1990

Thanks, but does the ln need to be in the Kraemer equation?

All I did was rearrange the equations you gave so you could get k' and k''

The k'' doesn't appear in the version of the Kraemer equation that has a ln in it.