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Standard XII

Chemistry

Dalton's Law of Partial Pressures

The value of ...

Question

The value of $K_{p}$ for the reaction
${C O}_{(2 (g)} + C_{(g)} ⇌ 2 {C O}_{(g)}$ is $3.0$ at $1000 K$ . If initially $P_{{C O}_{2}} = 0.48$ bar and $P_{C O} = 0$ bar and pure graphite is present. The equilibrium partial pressure of $C O$ and ${C O}_{2}$ are:

P_{C O} = 0.33, P_{{C O}_{2}} = 0.15

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P_{C O} = 0.66, P_{{C O}_{2}} = 0.33

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P_{C O} = 1.44, P_{{C O}_{2}} = 0.66

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P_{C O} = 0.66, P_{{C O}_{2}} = 0.15

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Solution

The correct option is D $P_{C O} = 0.66, P_{{C O}_{2}} = 0.15$
Given $K_{p}$ =3, $P_{C O_{2}}$ =0.48 bar, $P_{C O}$ =0 bar
${C O}_{2} (g) + C (s) ⟺ 2 C O$
$t = 0 {C O}_{2} = 0.48 C O = 0 t = t_{e q} {C O}_{2} = 0.48 - p C O = 2 p$
$K_{P} = \frac{(p_{c o})^{2}}{p_{c o}}$
$3 = \frac{(2 p)^{2}}{0.48 - p}$
p=0.3325 bar
$p_{c o} = 2 p = 2 \times 0.3325 b a r = 0.665 b a r$
$p_{c o_{2}} = 0.48 - p = 0.48 - 0.3352 b a r = 0.1475 b a r = 0.15 b a r$

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The value of Kp for the reactionCO(2(g)+C(g)⇌2CO(g) is 3.0 at 1000K. If initially PCO2=0.48 bar andPCO=0 bar and pure graphite is present. The equilibrium partial pressure of CO and CO2 are:

The correct option is D PCO=0.66,PCO2=0.15Given Kp=3, PCO2=0.48 bar, PCO=0 barCO2(g)+C(s)⟺2COt=0CO2=0.48CO=0t=teqCO2=0.48−pCO=2pKP=(pco)2pco3=(2p)20.48−pp=0.3325 barpco=2p=2×0.3325bar=0.665barpco2=0.48−p=0.48−0.3352bar=0.1475bar=0.15bar

The value of $K_{p}$ for the reaction
${C O}_{(2 (g)} + C_{(g)} ⇌ 2 {C O}_{(g)}$ is $3.0$ at $1000 K$ . If initially $P_{{C O}_{2}} = 0.48$ bar and $P_{C O} = 0$ bar and pure graphite is present. The equilibrium partial pressure of $C O$ and ${C O}_{2}$ are: