Question

For the reaction `"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)`, at what temperature does `K_c` numerically equal `K_p`?

`a)` `"0.1203 K"`
`b)` `"12.19 K"`
`c)` `"273 K"`
`d)` `"298 K"`

Answer 1

I got `"12.19 K"`.

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For ideal gases, `PV = nRT`. Note that:

• `n/V` is concentration in `"mol/L"`
• `P_("N"_2"O"_4) = (nRT)/(V) = ["N"_2"O"_4]RT`, the partial pressure of `"N"_2"O"_4`.
• `P_("NO"_2) = (nRT)/(V) = ["NO"_2]RT`, the partial pressure of `"NO"_2`.

Recall that stoichiometric coefficients become exponents in equilibrium expressions for gases and aqueous solutions.

So, to relate `K_c` and `K_p`, the expression would be:

`K_c = (["NO"_2]^2)/(["N"_2"O"_4])`

`bb(K_p) = (P_("NO"_2)^2)/(P_("N"_2"O"_4))`

`= (["NO"_2]RT)^2/(["N"_2"O"_4]RT)`

`= bb(RT(["NO"_2]^2)/(["N"_2"O"_4]) = K_cRT`, with units of pressure

Since we are wondering when numerically, `K_c = K_p` for this reaction, we use the expression above in bold to solve for `T` in `"K"`.

`=> RTK_c = K_p` in `"atm"`

Thus, `RT` must numerically equal `1` if `K_c` is to be equal to `K_p`.

`=> color(blue)(T = (K_p)/(RK_c))`

It is implied that `K_p` is typically reported in units of `"atm"` (only `K_p` at standard conditions does not have units). Therefore, we would use `R = "0.082057 L"cdot"atm/mol"cdot"K"` (as opposed to `R = "0.083145 L"cdot"bar/mol"cdot"K"`, or `"8.314472 J/mol"cdot"K"`).

So, `color(blue)(T) = 1/0.082057 * 1/1` `"K"` `~~` `color(blue)("12.19 K")`

Answer 2

`sf(12.19color(white)(x)K)`

Explanation:

The relationship between `sf(K_c)` and `sf(K_p)` is :

`sf(K_p=K_c(RT)^(Deltan))`

`sf(Deltan)` is the no. moles product - no. moles reactant

`sf(Deltan=2-1=1)`

The only way that `sf(K_p)` can have the same numerical value as `sf(K_c)` is when `sf((RT)^1=1)`

`:.sf(T=1/R=1/0.082=12.19color(white)(x)K)`

When the question states `sf(K_p=K_c)` it should state the dimensions that are being used. If pressure was in `sf("N/m"^2)` then the value of 8.31 J/K/mol should be used for R which gives a different value for T.

You need to assume that pressure is normalised against 1 atmosphere and concentration is normalised against unit concentration.

A badly worded question.