An industrial chemist puts 1.40 mol each of H2(g) and CO2(g) in a 1.00-L container at a certain temperature. When equilibrium is reached, 0.52 mol of CO(g) is in the container. Find Keq at this temperature for the following reaction: H2(g) + CO2(g)↔H2O(g) + CO(g)

H₂(g) + CO₂(g) ⇆ H₂O(g) + CO(g)             Kc → 0.35

Explanation:

The equilibrium is:

H₂(g) + CO₂(g) ⇆ H₂O(g) + CO(g)

Initially we have 1.40 moles of hydrogen and 1.40 mol of carbon dioxide.

During the reaction x amount has reacted.

In the equilibrium we have 0.52 moles of CO. Therefore the scheme is:

               H₂(g)  +   CO₂(g)   ⇆    H₂O(g)   +    CO(g)

Initially     1.40          1.40                 -                  -

React         x                x                   x                x

Eq.        1.40-x         1.40-x             0+x             0+x=0.52

Then x = 0.52. Ratio is 1:1, so If x reacted, x is formed during the equilibrium

1.40 - 0.52 = 0.88 → [H₂] and [CO₂]

0.52 = [CO] and [H₂O]

The expression for Kc is:  [CO] . [H₂O] / [H₂] . [CO₂]

Kc = (0.52 . 0.52) / (0.88 . 0.88) → 0.35

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.