What is ∆h0 for this reaction?

br2 (g) → br2 (l)

∆hf ° br2 (g) = 30.91 kj/mol

∆H° of reaction is -30.91 KJ/mol.

Explanation:

Given data:

∆Hf ° of Br2 (g) = 30.91 KJ/mol

∆H° of reaction = ?

Solution:

we know that standard heat of formation for Br2 (l) is 0.0 KJ/mol.

Balance chemical equation:

Br2 (g) → Br2 (l)

Formula:

∆H° = ∆Hf ° (product) - ∆Hf °(reactant)

∆H° = 0.0 KJ/mol - 30.91 KJ/mol

∆H° = - 30.91 KJ/mol

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

gas

explanation:

smoke is the matter of a gas