Question

SHOW WORK.

Part 1: Methane (CH₄) combusts as shown in the equation below. If 3 moles of methane were used in the combustion, what would be the change in enthalpy (ΔH) for the reaction?

CH_4(g) + 2 O_2(g) → CO_2(g) + 2 H₂O_(g) ΔH = −802 kJ

Part 2: The specific heat of water is 4.184 J/gK. How many joules (J) of heat are required to heat 15 g of water from 19.00 °C to 35.00 °C?

3. Find the heat of reaction for the following process:

Part 3: FeCl₂(s) → Fe(s) + 2 FeCl₃(s)

given that:

Fe(s) + 2HCl(aq) → FeCl₂(s) + H₂(g) ∆H = -769 kJ

2Fe(s) + 6HCl(aq) → 2FeCl₃(s) + 3H₂(g) ∆H = -1,822 kJ

If possible, can someone answer the three parts? If not just one of the parts is fine.

Answer 1

1)

From reaction,

When 1 mol of CH4 reacts, ΔH = -802 KJ

So,

For 3 mol of CH4 reacting, ΔH = -802*3 = -2406 KJ

Answer: -2406 KJ

2)

Given:

m = 15 g

C = 4.184 J/g.oC

Ti = 19 oC

Tf = 35 oC

use:

Q = m*C*(Tf-Ti)

Q = 15.0*4.184*(35.0-19.0)

Q = 1004 J

Answer: 1.00*10^3 J

3)

Lets number the reaction as 0, 1, 2 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 0 = -3 * (reaction 1) +1 * (reaction 2)

So, ΔHo rxn for required reaction will be:

ΔHo rxn = -3 * ΔHo rxn(reaction 1) +1 * ΔHo rxn(reaction 2)

= -3 * (-769) +1 * (-1822)

= 485 KJ

Answer: 485 KJ