If $K_p = 2.4 * 10^(-3)$ for the reaction below, then what is $K_c$ ?

Md Ariful Islam

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Md Ariful Islam

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Your tool of choice here will be the equation

$color(blue)(ul(color(black)(K_p = K_c * (RT)^(Deltan))))$

Here

  • $K_p$ is the equilibrium constant in terms of partial pressures
  • $K_c$ is the equilibrium constant in terms of concentrations
  • $R$ is the universal gas constant, equal to $0.0821 quad ("atm" * "L")/("mol" * "K")$
  • $T$ is the absolute temperature at which the reaction takes place
  • $Deltan$ is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants

Now, your reaction takes place at $1000^@"C"$, so start by converting the temperature from degrees Celsius to Kelvin.

$T = 1000^@"C" + 273.15 = "1273.15 K"$

Notice that for every $1$ mole of nitrogen gas that takes part in the reaction, the reaction consumes $3$ moles of hydrogen gas and produces $2$ moles of ammonia.

This means that you have

$Deltan = color(white)(overbrace(color(black)(" 2 "))^(color(blue)("moles of ammonia")) " "color(black)(-)" " overbrace(color(black)((" 1 + 3 ")))^(color(blue)("moles of reactants"))$

$Deltan = - 2$

Rearrange the equation to solve for $K_c$

$K_c = K_p/((RT)^(Deltan)$

Plug in your values to find--since you didn't provide any units for $K_p$, I'll do the calculation without added units!

$K_c = (2.4 * 10^(-3))/(0.0821 * 1273.15)^(-2) = color(darkgreen)(ul(color(black)(26)))$

The answer is rounded to two , the number of sig figs you have for $K_p$.

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