Initially at temperature of 90°C, 0.18mol/L CS2(g) and 0.31mol/L H2 are present in a closed container. At equilibrium 0.13mol/L CS2 is present how much H2 is present
To solve this problem, we can use the balanced chemical equation for the reaction between CS2 and H2:
CS2(g) + 2H2(g) ⇌ CH4(g) + H2S(g)
The stoichiometry of the reaction tells us that for every 1 mol of CS2 that reacts, 2 mol of H2 also react. Therefore, if 0.18 mol/L of CS2 reacts, then 2 * 0.18 = 0.36 mol/L of H2 also reacts.
Since we started with 0.31 mol/L of H2 initially, the amount of H2 present at equilibrium can be calculated by subtracting the amount of H2 that reacted from the initial amount:
0.31 mol/L - 0.36 mol/L = -0.05 mol/L
However, it is not possible to have a negative concentration of a substance. Therefore, the answer is that no H2 is present at equilibrium.