Question

You wish to find the enthalpy for the reaction 6 gei₄ (s) 14 nh₄i (s) → 3 ge₂h₆ (l) 7 n₂ (g) 38 hi (g) given the following equations equation 1: 2 ge (s) 3 h₂ (g) → ge₂h₆ (l) ∆h = 137.3 kj/mol equation 2: ge (s) 4 hi (g) → gei₄ (s) 2 h₂(g) ∆h = -247.8 kj/mol equation 3: 2 nh₄i (s) → n₂ (g) 2 hi (g) 3 h₂(g) ∆h = 455.8 kj/mol what would be the enthalpy change, in kj/mol, for 6 ge (s) 9 h₂ (g) → 3 ge₂h₆ (l)?

Answer 1

To find the enthalpy change for the reaction 6 Ge(s) + 9 H2(g) → 3 Ge2H6(l), we can use a combination of the given equations and the principles of Hess's law. The enthalpy change for the reaction is -457.2 kJ/mol.

Step-by-step explanation:

To find the enthalpy change for the reaction 6 Ge(s) + 9 H2(g) → 3 Ge2H6(l), we can use a combination of the given equations and the principles of Hess's law. First, we need to balance the equation with the given equations. By multiplying equation 1 by 3 and equation 2 by 4, we can cancel out Ge and H2 to get the desired equation.

Next, we can add the enthalpy changes of the individual reactions. Multiply equation 1 by 3 and equation 2 by 2 to get the desired equation. The enthalpy change for equation 1 is (+137.3 kJ/mol) * 3 and for equation 2 is (-247.8 kJ/mol) * 2.

Now, sum all the enthalpy changes: 3*(137.3) + 2*(-247.8) = -457.2 kJ/mol. Therefore, the enthalpy change for the reaction 6 Ge(s) + 9 H2(g) → 3 Ge2H6(l) is -457.2 kJ/mol.

Answer 2

Using Hess's law and given reactions, the enthalpy change for the reaction 6 Ge (s) + 9 H2 (g) → 3 Ge2H6 (l) is calculated to be 705.23 kJ.

Step-by-step explanation:

To calculate the enthalpy change (ΔH) for the reaction 6 Ge (s) + 9 H2 (g) → 3 Ge2H6 (l), we need to use Hess's law and manipulations of the given equations. First, we need to identify how many times each given equation must be used to obtain the desired reaction when added together. The first equation is used in its current form (× 3), the second equation is reversed and used (× 6), and the third equation is used (× 7).

Using Hess's law:

• 3 × Equation 1: 3(137.3 kJ/mol) = 411.9 kJ
• 6 × Equation 2 (reversed): 6(–247.8 kJ/mol) = -1486.8 kJ
• 7 × Equation 3: 7(455.8 kJ/mol) = 3190.6 kJ

Summing the enthalpy changes of the individual steps:

(411.9 kJ) + (-1486.8 kJ) + (3190.6 kJ) = 2115.7 kJ

We then divide this sum by the number of times the target reaction occurs, which is three:

× 3 Target Reaction: 2115.7 kJ / 3 = 705.23 kJ

Therefore, the enthalpy change for the reaction 6 Ge (s) + 9 H2 (g) → 3 Ge2H6 (l) is 705.23 kJ.