CH4 reacts with CCl4 to form CH2Cl2. The equilibrium constant, Kp, for this reaction is 9.52×10-2 at 350 K. CH4(g) + CCl4(g) 2CH2Cl2(g) The standard enthalpy change for this reaction (ΔH°) is 18.8 kJ/mol. (a) Predict the effect on the production of CH2Cl2 when the temperature of the equilibrium system is increased. (b) Use the van't Hoff equation to estimate the equilibrium constant for this reaction at 474 K. Kp at 474 K =

Heating favors increasing CH₂Cl₂ yield as K₂ = 0.517 is greater than K₁ = 0.0952.

Explanation:

For endothermic reactions, heating the reaction system functions to increase product yield while decreasing exothermic reaction yield.

Given the reaction CH₄(g) + CCl₄(g) ⇄ 2CH₂Cl₂2(g ;  Kp = 0.952 also equals Kc b/c ∑ΔV(Reactants) = ∑ΔV(Products) => Δn = 0 in Kp = Kc(RT)^Δn => Kp = Kc(RT)^0 = Kc(1) = Kc

Consider the given reaction as being a stabilized equilibrium under the case I condition and balanced upon a see-saw as follows ...

                            CH₄(g) + CCl₄(g) + 18.8Kj/mol ⇄ CH₂Cl₂(g)

                                                                   Δ

Generally, heating an endothermic reaction functions to shift reaction away from the reactants and toward the product side of the process and increase yield of the product of interest. The opposite effect occurs for an exothermic reaction.

This effect is easily visualized if one hypothetically considers that  heating the endothermic process adds weight to the side of the reaction having the ΔH value,  such as ...

Part B

ln(K₂/K₁) = ΔH/R(1/T₁ - 1//T₂) => ln(K₂) = ln(K₁) + ΔH/R(1/T₁ - 1//T₂)

ln(K₂) = ln(0.0952) + 18.8/(0.008314)(1/350 - 1/474) = -0.66

K₂ = e⁻⁰°⁶⁶ = 0.517 at 474K

K₁ = 0.0952 at 350K << K₂ = 0.517 at 474K => reaction shifts right to produce more product.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'