For the reaction a (g) → 2 b (g), k = 14.7 at 298 k. what is the value of q for this reaction at 298 k when ∆g = -20.5 kj/mol?

Q= 250= 2,5 x 10^2

Explanation:

ΔG=ΔGº +RTlnQ. R=8,314 J/molK and T=298K

In the balance ΔG=ΔGº +RTlnK=0 → ΔGº= -RTLnK= - 6,7 KJ/mol

ΔG=ΔGº +RTlnQ → -20,5 =  -6,7  + 2,5 LnQ

→LnQ= -5,5 → Q=250

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

is has three carbons, so the answer is propylamine.

meth-

eth-

prop-

but-

pent-

hex-

hept-

oct-

non-

explanation: