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love history [14]
3 years ago
11

When HCl(g) reacts with NH3(g) to form NH4Cl(s), 42.1 kcal of energy are evolved for each mole of HCl(g) that reacts. Write a ba

lanced equation for the reaction with an energy term in kcal as part of the equation.
Chemistry
1 answer:
stich3 [128]3 years ago
8 0

<u>Answer:</u> The chemical equation is HCl(g)+NH_3(g)\rightarrow NH_4Cl(s)+42.1kCal

<u>Explanation:</u>

There are 2 types of reactions that are classified based on enthalpy change:

  • Endothermic reaction
  • Exothermic reaction

Endothermic reactions: They are defined as the reactions where heat is absorbed by the reaction. The change in enthalpy of the reaction is always positive.

Exothermic reactions: They are defined as the reactions where heat is released by the reaction. The change in enthalpy of the reaction is always negative.

Given values:

Energy released for 1 mole of HCl reacted = -42.1 kCal

The chemical equation for the formation of ammonium chloride follows:

HCl(g)+NH_3(g)\rightarrow NH_4Cl(s)+42.1kCal

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ΔHf° is the standard heat of formation

The ΔHf° for the substances are the following:
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ΔHrxn° = (1 mol×-74.7 kJ/mol) - ∑(1 mol×-238.4 kJ/mol)
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2 years ago
You make a solution by putting 45.6g of iron lll carbonate into 167ml of water. What is it's molarity?
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3 years ago
The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl
AlekseyPX

Answer: Concentration of NH_3 in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of H_2 = 0.729 M

The given balanced equilibrium reaction is,

                 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

Initial conc.            x                0           0

At eqm. conc.     (x-2y) M     (y) M   (3y) M

The expression for equilibrium constant for this reaction will be:

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K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}

Now put all the given values in this expression, we get :

K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

x=0.80

concentration of NH_3 in the equilibrium mixture = 0.80-2\times 0.243=0.31

Thus concentration of NH_3 in the equilibrium mixture is 0.31 M

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