Answer:
-1,103.39KJ/mol
Explanation:
We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.
In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.
The standard enthalpies of the molecules above are as follows:
H2S = -20.63KJ/mol
H2O = -285.8KJ/mol
SO2 = -296.84KJ/mol
O2 = 0KJ/mol
ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3
ΔfH⦵(O2)]
ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]
-[ 3 × -20.63)]
= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
<span>Start with the number of grams of each element, given in the problem.Convert the mass of each element to moles using the molar mass from the periodic table.Divide each mole value by the smallest number of moles calculated.<span>Round to the nearest whole number.</span></span>
Answer:
It would increase the final quantity of products
Explanation:
According to the Le- Chatelier principle,
At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.
The equilibrium can be disturb,
By changing the concentration
By changing the volume
By changing the pressure
By changing the temperature
Consider the following chemical reaction.
Chemical reaction:
2NO₂ ⇄ N₂O₄
In this reaction the equilibrium is disturb by increasing the concentration of reactant.
When the concentration of reactant is increased the system will proceed in forward direction in order to regain the equilibrium. Because when reactant concentration is high it means reaction is not on equilibrium state. As the concentration of NO₂ increased the reaction proceed in forward direction to regain the equilibrium state and more product is formed.