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Novay_Z [31]
4 years ago
11

Which of the following is true about a system at equilibrium? a The concentration(s) of the reactant(s) is equal to the concentr

ation(s) of the product(s). b No new product molecules are formed. c The concentration(s) of reactant(s) is constant over time. d The rate of the reverse reaction is equal to the rate of the forward reaction and both rates are equal to zero. e None of the above (a-d) is true.
Chemistry
1 answer:
anyanavicka [17]4 years ago
6 0

Answer:

c The concentration(s) of reactant(s) is constant over time.  

Step-by-step explanation:

When the reaction A ⇌ B reaches equilibrium, the concentrations of reactants and products are constant over time.

a is <em>wrong</em>, because the concentrations of reactants and products are usually quite different.

b is <em>wrong</em>, because both product and reactant molecules are being formed at equilibrium.

d is <em>wrong</em>. The rates of the forward and reverse reactions are equal, but they are not zero.

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Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M

The given equilibrium reaction is,

                           N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially conc.      0        0           0.1576

At eqm.               (x)       (x)        (0.1576-2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}

By solving the term, we get:

x=0.0742,0.0839

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

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