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wolverine [178]
3 years ago
9

Consider the reaction: 2clf3(g) + 2nh3(g) → n2(g) + 6hf(g) + cl2(g) when calculating the δh°rxn, why is the δhf° for n2 not impo

rtant?
Chemistry
1 answer:
Leni [432]3 years ago
8 0

<span>Actually, the heat of reaction hrxn s calculated by taking the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. However, at heat of formations of pure elements at atmospheric conditions is zero, therefore the hf of N2 is not important since it is zero anyway.</span>

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.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f
Tasya [4]

Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

6 0
3 years ago
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What is the pH of 0.0001 M NaOH
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During a laboratory experiment, you discover that an enzyme-catalyzed reaction has a ∆G of -20 kcal/mol. If you double the amoun
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Answer:

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The dG can be expressed in terms of the G(products) - G(reactants). If the amount of enzyme is doubled the Gibbs energy of the reactants and products will be the same, so the substraction dG has the same value

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Carbon dioxide isn't a polar molecule because

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