Question

QUESTION 1 0.5 [CLO-4] The expression for Kp for the reaction below is 4CuO (s) + CH4 (8) CO2 (g) + 4Cu (s) + 2H20 (8) PcO2 PH202 PCH4 PCH4 [Cu] PCO, PH202 A) B) D) PCuo PCO2 PH22 PH2O2 PcO2 (CuOj4 PCH4 ===== PCO, PH202 E) PCH4 СА E D с B Save All Ar Click Save and Submit to save and submit. Click Save All Answers to save all answers.

Answer 1

Short answer: E

A chemical reaction is said to be in a state of equilbirum when the compostions of the reaction mixtures remain constant over a period of time; i.e., the net rate of formation of products is equal to its consumption in the reverse reaction.

At a state of equilibrium, a constant defined as the equlibrium constant can be used to determine the composition of the reaction mixture. It may be expressed as the ratio of the partial pressures of the reacting components (Kp) or as the ratio of the conentrations of the reacting components,

Kp, may be defined as:

Kp= product of partial pressures of gaseous products/ product of partial pressures of gaseous reactants; with each partial pressure raised to a power equal to the corresponding stoichiometric coefficient in the balanced chemical equation for the reaction.

The partial pressure of a gas in a mixture of gases is defined as the pressure exerted by the gas if it occupied the entire gaseous volume on its own. Please note that the partial pressures of pure solids and liquids is zero.

In the given question, the reaction procees as the following balanced chemical equation:

4 CuO_(s) + CH4 _(g) $\rightarrow$ CO_2(g) + 4 Cu_(s) + 2H₂O_(g)

Here, the partial pressures of CuO and Cu is equal to zero as both are solids.

Thus, Kp is given as:

Kp = P_CO2 PH₂O²/PCH4