A mixture of nitrogen gas, hydrogen gas, and ammonia gas are...

A mixture of nitrogen gas, hydrogen gas, and ammonia gas are allowed to come to equilibrium at $200^{\circ} \mathrm{C}$, and their partial pressures are found to be 0.61 atm $\mathrm{N}_2, 1.1 \mathrm{~atm} \mathrm{H}_2$, and $0.52 \mathrm{~atm} \mathrm{NH}_3$. a. Based on this data, what is $\mathrm{K}_P$ for the following equation? $$ \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) $$ b. Based on this data, what is $\mathrm{K}_P$ for the equation below? $$ 1 / 2 \mathrm{~N}_2(\mathrm{~g})+3 / 2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g}) $$

A mixture of nitrogen gas, hydrogen gas, and ammonia gas are allowed to come to equilibrium at $200^{\circ} \mathrm{C}$, and their partial pressures are found to be 0.61 atm $\mathrm{N}_2, 1.1 \mathrm{~atm} \mathrm{H}_2$, and $0.52 \mathrm{~atm} \mathrm{NH}_3$.
a. Based on this data, what is $\mathrm{K}_P$ for the following equation?

$$
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})
$$

b. Based on this data, what is $\mathrm{K}_P$ for the equation below?

$$
1 / 2 \mathrm{~N}_2(\mathrm{~g})+3 / 2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})
$$

Key Concepts

Equilibrium Constant in Terms of Partial Pressures (Kp)

This concept refers to the ratio of the products of the partial pressures of the gaseous products to those of the gaseous reactants, each raised to the power of their stoichiometric coefficients, at equilibrium. It provides a quantitative measure of the position of equilibrium for reactions involving gases and is an essential tool in predicting how changes in conditions affect the reaction.

Reaction Stoichiometry

Reaction stoichiometry involves the balanced coefficients in a chemical equation, dictating the proportionality between reactants and products. Accurate stoichiometry is critical because it determines the exponents in the equilibrium expression (whether in Kp or Kc), affecting how the individual partial pressures or concentrations contribute to the overall equilibrium constant.

Scaling of the Equilibrium Constant

When the entire chemical equation is multiplied or divided by a factor, the equilibrium constant is adjusted by raising it to the power of that factor. This is due to the relationship between the coefficients of the reaction and the exponents in the equilibrium expression, and it ensures that the equilibrium constant remains dimensionally and mathematically consistent with the modified reaction equation.

Transcript

00:01 We are given this reaction and we are told that it has an equilibrium constant kc value of 1 .2.

00:07 In part a, we want to determine what the value is of kp for this reaction.

00:13 We know that kp is equal to kc times rt to the power of delta and gas.

00:24 So we can calculate kp from this equation using the given value of kc 1 .2.

00:32 R is the ideal gas constant 0 .0 .0 .81 liters times atmospheres per mole times kelvin.

00:48 When we convert the temperature from celsius into kelvin, it comes out to 648.

00:55 Then we raise this to the power of delta n gas.

00:59 We see that we have two moles of gas on the product side and a total of four moles of gas on the reactant side.

01:05 So 2 minus 4 is negative 2.

01:07 So delta n gas is negative 2.

01:11 So we raise rt to the power of negative 2 and multiply it by 1 .2 to get a final answer of kp to be about 4 .2 times 10 to the power of negative 4.

01:33 And now in part b, we want to calculate the value of kc for this reaction.

01:39 If we compare it with the reaction that we are given, we see that the reaction in part b is the reverse of that reaction.

01:48 We had n2 in 3h2 gas as reactants.

01:53 Now we have n2 and 3h2 gas as products.

01:56 We had 2nh3 as a product and now we have 2nh3 as a reactant.

02:04 So all we did was reverse that reaction.

02:10 And so when we reverse a reaction with a known value of kc, we take the inverse.

02:14 To get the kc of the reverse of that reaction.

02:19 So kc for this reaction is equal to 1 over kc for the forward reaction, which was 1 .2...

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