Question

Chapter 15 (15.87)

At 700 K the equilibrium constant for the reaction
CCl4(g)←−→C(s)+2Cl2(g)
is Kp=0.76. A flask is charged with 2.00 atm of CCl4, which then reaches equilibrium at 700 K.

1) What fraction of the CCl4 is converted into C and Cl2?

2) What is the partial pressure of CCl4 at equilibrium?

3) What is the partial pressure of Cl2 at equilibrium?

Answer 1

Answer on Question #63428, Chemistry / General Chemistry

Chapter 15 (15.87)

At 700 K the equilibrium constant for the reaction

$$CCl_4(g) \leftarrow \rightarrow C(s) + 2Cl_2(g)$$

is $K_p = 0.76$. A flask is charged with 2.00 atm of $CCl_4$, which then reaches equilibrium at 700 K.

1) What fraction of the $CCl_4$ is converted into C and $Cl_2$?

2) What is the partial pressure of $CCl_4$ at equilibrium?

3) What is the partial pressure of $Cl_2$ at equilibrium?

Solution:

$$CCl_4(g) \leftarrow \rightarrow C(s) + 2Cl_2(g)$$ $$pCCl_4 \dots pCl_2$$

initial...2.00...0

change...-x...+2x

equil...2.00-x...2x

$$K_p = \frac{pCl_2}{2} \cdot 2 / pCCl_4$$ $$0.76 = \frac{2x}{2} \cdot 2 / 2.00 - x$$ $$1.52 - 0.76x = 4x^2$$ $$x = 0.529$$

Change in pressure of $CCl_4 = 0.529$ atm

1) 0.529 atm / 2.00 atm = 0.265

2) $pCCl_4 = 2.00 - 0.529 = 1.471$ atm

3) $pCl_2 = 2(0.529) = 1.058$ atm

Answer: 1) 0.265; 2) 1.471 atm; 3) 1.058 atm

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