[ashyyamora]

Newcomer to the forum here, so I'm not sure if I'm posting this in the right place nor in the right format, so please correct me on that.

A question found on an online Chemistry 12 exam was:
Consider the following equilibrium system:
C_(s) + 2H₂ CH_4(g) + 75kJ
State three different ways to make more C_(s) react.

The answers were:
Any three of the following:
• add H2
• remove CH4
• decrease temperature
• increase pressure/decrease volume

I was confused though because doesn't "More C_(s) reacting" just mean that the reaction rate of C_(s) is higher and more solid carbon is being used, regardless of how much solid carbon is being produced? Wouldn't decreasing the temperature actually decrease the rate of all the chemical species, including C_(s), and therefore mean less C_(s) is reacting? And would things like adding a catalyst, or increasing the surface area of the solid carbon also increase the reaction rate of C_(s), making more C_(s) react, regardless of the fact that just as much C_(s) is being produced?

The explanation I was given is that "make more C_(s) react" means the same thing as shifting to the right. However, shifting to the right just means more C_(s) is reacting than being produced, not necessarily that more C_(s) is reacting than it was before, so I don't understand the answer.

Thanks guys.

[Borek]

You are mistaking kinetics for equilibrium. Yes, lower temperature means slower reactions, but how much C(s) (note: (s) should not be a subscript) reacts in the end depends not on how fast the reaction is, but where the equilibrium lies.

[XeLa.]

note: (s) should not be a subscript

Some textbooks do list state symbols as subscripts, but it is more convenient and easier to read when state symbols are written as (s).

[Borek]

note: (s) should not be a subscript

Some textbooks do list state symbols as subscripts, but it is more convenient and easier to read when state symbols are written as (s).

Books can do whatever they want, IUPAC clearly defines state symbols as a normal script.

Compare http://www.chemicalforums.com/index.php?topic=8373.msg38043#msg38043

[P]

If the carbon is solid...  then the more finely powdered it is the faster it will react. Won't make more of it react though. Is there a catalyst that promotes further reaction?...

Removal of the product should make the reaction continue. I do not know what answer they actually want though.