# Decomposition of C5H6O3 equilibrium

• Chemistry
• i_love_science
So your final answer should be##K_c = \dfrac{0.389 \times \mathrm{L^3\;atm^{-3}}}{(0.08206 \; \mathrm{L\; atm\;mol^{-1} \; K^{-1}})^3} = 6.74 \times 10^{-6} \; \mathrm{mol^{-1} \; L^3}##.In summary, after calculating the partial pressure and equilibrium partial pressure of each component, the total pressure was found to be 1.63 atm. However, the resulting value of Kp was incorrect due to an error in converting to Kc. The correct value of Kc is

#### i_love_science

Homework Statement
Consider the decomposition of the compound C5H6O3 as follows:
C5H6O3(g) --> C2H6(g) + 3CO(g)
When a 5.63-g sample of pure C5H6O3(g) was sealed in an otherwise empty 2.50-L flask and heated to 200°C, the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate Kc for this reaction.
Relevant Equations
equilibrium expression
My solution:

partial pressure of C5H6O3 = mRT/MV = (5.63 g)(0.08206 L*atm/mol*K)(473 K) / (114.098 g/mol)(2.50 L) = 0.766 atm

equilibrium partial pressure of C5H6O3 = 0.766 - x
equilibrium partial pressure of C2H6 = x
equilibrium partial pressure of CO = 3x
total pressure = 0.766 atm - x + x + 3x = 1.63 atm
x = 0.288 atm
Kp = (0.288 atm)(3*0.288 atm)^3 / (0.766-0.288) = 0.389
Kc = Kp / (RT)^(delta n) = 0.389 / [(0.08206 L*atm/mol*K)(473 K)]^1 = 0.01

My answer was wrong. The correct answer is 6.74*10^-6, and was found by doing all the calculations directly in terms of moles and concentration. Could anyone explain where I went wrong? Thanks.

Nearly all fine, except for the final conversion from ##K_p## to ##K_c##. Notice that\begin{align*}

K_c = \dfrac{[\mathrm{CO}]^3 [\mathrm{C_2H_6}]}{[\mathrm{C_5 H_6 O_3}]} &= \dfrac{\left(\dfrac{n_{\mathrm{CO}}}{V}\right)^3 \left(\dfrac{n_{\mathrm{C_2H_6}}}{V}\right)}{\left(\dfrac{n_{\mathrm{C_5 H_6 O_3}}}{V}\right)} \times \dfrac{(RT)^4}{(RT)^4} \\ \\

&= \dfrac{\left(\dfrac{n_{\mathrm{CO}} RT }{V}\right)^3 \left(\dfrac{n_{\mathrm{C_2H_6}} RT}{V}\right)}{\left(\dfrac{n_{\mathrm{C_5 H_6 O_3}} RT}{V}\right)} \times \dfrac{1}{(RT)^3} = \dfrac{K_p}{(RT)^3}

\end{align*}In other words, in the formula ##K_c = K_p / (RT)^{\Delta n}## you should have ##\Delta n = 3##, which you can tell directly from the reaction equation.

i_love_science

## 1. What is the chemical equation for the decomposition of C5H6O3?

The chemical equation for the decomposition of C5H6O3 is C5H6O3 → 3CO + 3H2.

## 2. What factors can affect the equilibrium of the decomposition of C5H6O3?

The equilibrium of the decomposition of C5H6O3 can be affected by temperature, pressure, and the presence of catalysts.

## 3. What is the role of temperature in the decomposition of C5H6O3 equilibrium?

Temperature plays a crucial role in the decomposition of C5H6O3 equilibrium. An increase in temperature can shift the equilibrium towards the products, while a decrease in temperature can shift it towards the reactants.

## 4. How does the presence of a catalyst affect the decomposition of C5H6O3 equilibrium?

A catalyst can speed up the decomposition of C5H6O3 by lowering the activation energy required for the reaction to occur. This allows the reaction to reach equilibrium faster.

## 5. What is the significance of the decomposition of C5H6O3 equilibrium in real-world applications?

The decomposition of C5H6O3 equilibrium is significant in various industries, such as the production of polymers, plastics, and pharmaceuticals. It is also used in the production of fuels and as a method for waste management.