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velikii [3]
3 years ago
9

In certain conditions, the equilibrium concentrations for components in the reaction 2NOCl(g) 2NO(g) + Cl2(g) are [NO] = 0.02 M,

[Cl2] = 0.01 M, and [NOCl] = 0.5 M. What is the value of Keq, given Keq =?
Chemistry
1 answer:
Sonja [21]3 years ago
8 0
Keq= (products)/ (reactants)

Keq= ( [NO]^2 x [Cl2]) / ( [NOCl]^2)

Keq= ( (0.02)^2 x (0.01) ) / (0.5)^2= 1.6 x 10-5
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vitfil [10]

Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]

We are given:

\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo

Putting values in above equation, we get:

\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

6 0
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Answer:

C

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2 years ago
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