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MariettaO [177]
3 years ago
15

Writing a pressure equilibrium constant expression Ammonia and oxygen react to form nitrogen monoxide and water, like this: 4 NH

3(g)+5 O2(g)-4 NO(g)+6 H2O(g) Write the pressure equilibrium constant expression for this reaction.
Chemistry
1 answer:
Sliva [168]3 years ago
5 0

<u>Answer:</u> The equilibrium constant expression for the given equation is written below.

<u>Explanation:</u>

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{p}

For the general chemical equation:

aA+bB\rightleftharpoons cC+dD

The expression of K_p follows:

K_p=\frac{p_C^c\times p_D^d}{p_A^a\times p_B^b}

For the given chemical equation:

4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

The expression of K_p for above equation follows:

K_p=\frac{(p_{H_2O}^5)\times (p_{NO}^4)}{(p_{NH_3}^4)\times (p_{O_2}^5)}

Hence, the equilibrium constant expression for the given equation is written above.

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PLEASE HELP ME ASAP PLEASE!!!
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Answer:

\large \boxed{\text{1763 psi}}

Explanation:

We can use Dalton's Law of Partial Pressures:

Each gas in a mixture of gases equals its pressure independently of the other gases

\begin{array}{rcl}p_{\text{NO2}} + p_{\text{CO2}} & = & p_{\text{tot}} \\p_{\text{NO2}} + \text{795 psi} & = &\text{2558 psi}  \\p_{\text{NO2}} & = &\text{2558 psi - 795 psi} \\& = & \textbf{1763 psi}\\\end{array}\\\text{The partial pressure of nitrogen dioxide is $\large \boxed{\textbf{1763 psi}}$}

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Given the problem below, what would the first step be in solving it?
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Answer:

First step would be convert to moles

Final Answer: 37.8 g of NaCl

Explanation:

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We convert the mass of each reactant to moles:

18 g . 1mol /23g = 0.783 moles of Na

23g . 1mol / 70.9g = 0.324 moles of chlorine

We use the mole ratio to determine the limiting reactant:

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Then, 0.783 moles of Na, may react to (0.783 . 1)/2 = 0.391 moles.

Excellent!. We need 0.391 moles of Cl₂ and we only have 0.324 moles available. That's why the Cl₂ is our limiting reactant.

We use the mole ratio again, with the product side. (1:2)

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