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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

Assume that t...

Question

Assume that the decomposition of $H N O_{3}$ can be represented by the following equation:
$4 H N O_{3} (g) \to 4 N O_{2} (g) + 2 H_{2} O (g) + O_{2} (g)$
and the reaction approaches equilibrium at 400K temperature and 30atm pressure equilibrium partial pressure of $H N O_{3}$ is 2 atm.
calculate $K_{C}$ in $(m o l / L)^{3}$ at 400K:

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Solution

The correct option is D 32
$P_{t o t a l} = P_{H N O_{3}} + P_{N O_{2}} + P_{H_{2} O} + P_{O_{2}}$
$∵ P_{N O_{2}} = 4 P_{O_{2}}$ and $P_{H_{2} O} = 2 P_{O_{2}}$
$∴ P_{t o t a l} = P_{H N O_{3}} + 7 P_{O_{2}}$
$\Rightarrow 30 - 2 = P_{O_{2}} \times 7$
$\Rightarrow P_{O_{2}} = \frac{28}{7} = 4$
$K_{p} = \frac{P_{N O_{2}}^{2} . P_{H_{2} O} . P_{O_{2}}}{P_{H N O_{3}}^{4}}$
$= \frac{(4 \times 4)^{4} \times (2 \times 4)^{2} \times 4}{2^{4}} = 2^{20}$
$K_{p} = K_{c} (R T)^{Δ n_{g}} = K_{c} (0.08 \times 400)^{3}$
$\Rightarrow K_{c} = \frac{20^{20}}{(32)^{3}} = 32$

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Similar questions

Q. Assume that decomposition of

H N O_{3}

can be represented by the following equation

4 H N O_{3} (g) ⇌ 4 N O_{2} (g) + 2 H_{2} O (g) + O_{2} (g)

and the reaction approaches equilibrium at

400 K

temperature and

30 a t m

pressure. At equilibrium partial pressure of

H N O_{3}

2 a t m

.
Calculate

K_{c}

{(m o l / L)}^{3}

400 K

:(Use :

R = 0.08 a t m - L / m o l - K

)

Q. Assume that the decomposition of

H N O_{3}

can be represented by the following equation

4 H N O_{3} (g) ⇌ 4 N O_{2} (g) + 2 H_{2} O (g) + O_{2} (g)

and the reaction approaches equilibrium at

400 K

temperature and 30 atm pressure. At equilibrium partial pressure of

H N O_{3}

2 a t m

. Calculate

K_{c}

400 K

:
(Use R= 0.08 Latm/molK)

Q. Assume that the decomposition of

H N O_{3}

can be represented by the following equation:

4 H N O_{3} (g) ⇌ 4 N O_{2} (g) + 2 H_{2} O (g) + O_{2} (g)

.
The reaction approaches equilibrium at 400 K temperature and 30 atm pressure. At equilibrium, the partial pressure of

H N O_{3}

is 2 atm.
The value of

K_{c}

(

(m o l / L)^{3}

) at 400 K is:(Use

: R = .08 a t m - L / m o l e - K

)

For thedecomposition reaction 4 $H N O_{3}$ (g) $⇋$ 4 $N O_{2}$ (g) + 2 $H_{2}$ O (g) + $O_{2}$ (g) if we start with only the reactant $H N O_{3}$ and p = equilibrium pressure, the equilibrium constant $K_{p}$ is:

For the decomposition reaction $4 H N O_{3} (g) ⇌ 4 N O_{2} (g) + 2 H_{2} O + O_{2}$ (g) if we start with only the reactant $H N O_{3}$ and p = equilibrium pressure, the equilibrium constant $K_{p}$ is:

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Assume that the decomposition of HNO3 can be represented by the following equation:4HNO3(g)→4NO2(g)+2H2O(g)+O2(g)and the reaction approaches equilibrium at 400K temperature and 30atm pressure equilibrium partial pressure of HNO3 is 2 atm.calculate KC in (mol/L)3 at 400K:

The correct option is D 32Ptotal=PHNO3+PNO2+PH2O+PO2∵PNO2=4PO2 and PH2O=2PO2∴Ptotal=PHNO3+7PO2⇒30−2=PO2×7⇒PO2=287=4Kp=P2NO2.PH2O.PO2P4HNO3=(4×4)4×(2×4)2×424=220Kp=Kc(RT)Δng=Kc(0.08×400)3⇒Kc=2020(32)3=32