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CCI4(g) dissociates 50% into gaseous free radicals at 1000 K and at equilibrium experimental vapour density is found to be 38.5. Thus, CCI4(g) dissociates under the given conditions to
- a)CCl3(g), Cl(g)
- b)CCl2(g), Cl(g)
- c)C(g),CI(g)
- d)CCI(g),CI(g
Correct answer is option 'B'. Can you explain this answer?
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CCI4(g) dissociates 50% into gaseous free radicals at 1000 K and at eq...
Given information:
- CCI4(g) dissociates 50% into gaseous free radicals at 1000 K
- at equilibrium experimental vapour density is found to be 38.5
To determine the products of dissociation, we need to write the balanced chemical equation for the dissociation of CCI4:
CCI4(g) ⇌ CCl2(g) + Cl2(g)
The equilibrium constant for this reaction can be expressed as:
Kc = [CCl2][Cl2] / [CCI4]
Since we know the experimental vapour density of CCI4 at equilibrium, we can calculate its molar concentration:
n/V = P/RT
n/V = 38.5 / (0.0821 x 1000)
n/V = 0.467 mol/L
Assuming complete dissociation of CCI4, the initial concentration of CCI4 would be 0.2345 mol/L. Since the dissociation is 50%, the equilibrium concentration of CCI4 would be half of the initial concentration, i.e. 0.11725 mol/L.
Now we can use the equilibrium constant expression to calculate the concentrations of CCl2 and Cl2 at equilibrium:
Kc = [CCl2][Cl2] / [CCI4]
0.5 = [x][x] / (0.11725 - x)
where x is the equilibrium concentration of CCl2 and Cl2 (since they are formed in a 1:1 ratio)
Solving for x gives x = 0.170 mol/L. Therefore, the equilibrium concentrations of CCl2 and Cl2 are both 0.170 mol/L.
Thus, the products of dissociation are CCl2(g) and Cl2(g), which corresponds to option B.
- CCI4(g) dissociates 50% into gaseous free radicals at 1000 K
- at equilibrium experimental vapour density is found to be 38.5
To determine the products of dissociation, we need to write the balanced chemical equation for the dissociation of CCI4:
CCI4(g) ⇌ CCl2(g) + Cl2(g)
The equilibrium constant for this reaction can be expressed as:
Kc = [CCl2][Cl2] / [CCI4]
Since we know the experimental vapour density of CCI4 at equilibrium, we can calculate its molar concentration:
n/V = P/RT
n/V = 38.5 / (0.0821 x 1000)
n/V = 0.467 mol/L
Assuming complete dissociation of CCI4, the initial concentration of CCI4 would be 0.2345 mol/L. Since the dissociation is 50%, the equilibrium concentration of CCI4 would be half of the initial concentration, i.e. 0.11725 mol/L.
Now we can use the equilibrium constant expression to calculate the concentrations of CCl2 and Cl2 at equilibrium:
Kc = [CCl2][Cl2] / [CCI4]
0.5 = [x][x] / (0.11725 - x)
where x is the equilibrium concentration of CCl2 and Cl2 (since they are formed in a 1:1 ratio)
Solving for x gives x = 0.170 mol/L. Therefore, the equilibrium concentrations of CCl2 and Cl2 are both 0.170 mol/L.
Thus, the products of dissociation are CCl2(g) and Cl2(g), which corresponds to option B.
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CCI4(g) dissociates 50% into gaseous free radicals at 1000 K and at equilibrium experimental vapour density is found to be 38.5. Thus, CCI4(g) dissociates under the given conditions toa)CCl3(g), Cl(g)b)CCl2(g), Cl(g)c)C(g),CI(g)d)CCI(g),CI(gCorrect answer is option 'B'. Can you explain this answer?
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