When 1.50 mol of CH₄(g) reacts with excess Cl₂(g) at constant pressure according to the chemical equation shown below, 1062 kJ of heat are released. Calculate the value of ΔH for this reaction, as written.
2CH₄(g) + 3Cl₂(g) → 2CHCl₃(l) + 3H₂(g) H° = ?

Question

When 1.50 mol of CH₄(g) reacts with excess Cl₂(g) at constant pressure according to the chemical equation shown below, 1062 kJ of heat are released. Calculate the value of ΔH for this reaction, as written.
2CH₄(g) + 3Cl₂(g) → 2CHCl₃(l) + 3H₂(g) When 1.50 mol of CH4(g) reacts with excess Cl2(g) at constant pressure according to the chemical equation shown below, 1062 kJ of heat are released. Calculate the value of ΔH for this reaction, as written. 2CH4(g) + 3Cl2(g) → 2CHCl3(l) + 3H2(g) H° = ? A) -1420 kJ B) -708 kJ C) +708 kJ D) +1420 kJ

When 1.50 mol of CH<sub>4</sub>(g) reacts with excess Cl<sub>2</sub>(g) at constant pressure according to the chemical equation shown below, 1062 kJ of heat are released. Calculate the value of ΔH for this reaction, as written. 2CH<sub>4</sub>(g) + 3Cl<sub>2</sub>(g) → 2CHCl<sub>3</sub>(l) + 3H<sub>2</sub>(g) H° = ? A) -1420 kJ B) -708 kJ C) +708 kJ D) +1420 kJ

H° = ?

Quizplus · Accepted Answer

The given quantity of heat released in the reaction is -1062 kJ. This represents the value of ΔH for the reaction, since it is measured at constant pressure. The negative sign indicates that the reaction is exothermic, which means that energy is being released from the system to the surroundings. To obtain the value of ΔH in kJ/mol, we need to divide the given value by the number of moles of the limiting reactant, which in this case is 1.5 mol of CH₄. ΔH = -1062 kJ ÷ 1.5 mol = -708 kJ/mol The correct choice is A, which is the value of ΔH in kJ/mol, rounded to the nearest whole number.

When 150 Mol of CH4(g) Reacts with Excess Cl2(g) at Constant

When 150 Mol of CH₄(g) Reacts with Excess Cl₂(g) at Constant