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For the reaction N(2)O(4)(g)hArr2NO(2)(g...

For the reaction `N_(2)O_(4)(g)hArr2NO_(2)(g)`, the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?

A

The reaction is endothermic

B

The reaction is exothermic

C

If `NO_(2) (g) and N_(2) O_(4)(g)` are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more `N_(2)O_(4) (g)` will be formed,

D

The entropy of the system increases.

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
(a) As the value of K increases with increase of temperature and `K=(k_(f))/(k_(b))`, this means that `k_(f)` increases, i.e., formard reaction is favoured. Hence, reaction is endothermic.
(c) At 400 K, `Q=(p_(NO_(2))^(2))/(p_(N_(2)O_(4)))=((20)^(2))/(2) = 200 `. Thus `Q gt K`. Equilibrium will shift backward to form more `N_(2)O_(4)`.
(d) As reaction is accompanied by increase in the number of moles , entropy increases.
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Knowledge Check

  • For the reaction, N_(2)O_(4(g))iff2NO_(2(g)) , the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is/are correct?

    A
    The reaction is endothermic
    B
    The reaction is exothermic
    C
    If `NO_(2(g))andN_(2)O_(4(g))` are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more `N_(2)O_(4(g))` will be formed
    D
    The entropy of the system remains constant

  • For the reaction N_(2)O_(4)(g)hArr2NO_(2)(g) , the value of K_(p) is 1.7xx10^(3) at 500K and 1.7xx10^(4) at 600K . Which of the following is/are correct ?

    A
    The proportions of `NO_(2)` in the equilibrium mixture is increased by decrease in pressure.
    B
    The standard enthalpy change for the forward reaction is negative
    C
    Units of `K_(p)` are `atm^(-1)`
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  • For the reaction CO(g)+(1)/(2) O_(2)(g) hArr CO_(2)(g),K_(p)//K_(c) is

    A
    RT
    B
    `(RT)^(-1)`
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    D
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