An indication that an equillibrium system favors the reactants is a large K_eq.

Question

An indication that an equillibrium system favors the reactants is a large K_eq. Positive deltaH. Small K_eq. One step mechanism. Low activation energy. Ethene, C_2H_4, can be produced in the following industrial system: C_2H_6_(g) + energy rightleftarrows C_2H_4_(g) + H_2_(g) The conditions that are necessary to maximize the equillibrium yield of C_2H_4 are low temperater and low pressure. low temperater and high pressure. high temperater and low pressure. high temperater and high pressure. Consider the following equillibrium: 2NO_(g) + Br_2_(g) + energy rightleftarrows 2NOBr_(g) The equillibrium will shift to the left as a result of adding a catalyst. removing NOBr. increasing the volume. increasing the temperature. Consider the following equillibrium: CO_(g) + H_2O_(g) rightleftarrows CO_2_(g) + H_2_(g) deltaH = -41kL What will cause a shift to the left? adding a catalyst decreasing pressure by increasing volume adding CO increasing the temperature decreasing the temperature.

Explanation / Answer

4.

reactants --> left side of reaction

a large K will sate K = products/reactants, tha tis, high products

dH has nothing to do with shift of equilibrium

A small K = product / reactants, states that reactants > products, choose this option

1 step mechanism has nothing to do with equlibrium

a low activation energy has nothing to do with the equilibrium

therefore

only K < 1 (a Keq very low) will favour reactants

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