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tia_tia [17]
3 years ago
5

Consider the following reaction: Consider the reaction 2NO(g)+Br2(g)⇌2NOBr(g) ,Kp=28.4 at 298 K In a reaction mixture at equilib

rium, the partial pressure of NO is 119 torr and that of Br2 is 151 torr . What is the partial pressure of NOBr in this mixture?
Chemistry
1 answer:
Vitek1552 [10]3 years ago
4 0

Answer:  partial pressure of NOBr is 7792 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant is given as:

K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}

28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}

[p_{NOBr}]=7792 atm

Partial pressure of NOBr is 7792 atm

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How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

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