Pls answer this asap

1. The enthalpy change for the backward reaction is –20 KJmol¯¹

2. The forward reaction is endothermic.

3. The activation energy for the forward reaction is +70 KJmol¯¹.

Explanation:

Considering the diagram given above, the following data were obtained:

1. The enthalpy change for the forward reaction being is +20 KJmol¯¹ and so, the enthalpy change for the backward reaction is –20 KJmol¯¹

2. The heat energy of the product is higher than that of the reactant. Therefore, the forward reaction is endothermic.

3. Activation energy is the minimum energy needed to be overcome in order for the reaction to proceed to product. It can be obtained from the energy profile diagram by considering the energy between the reactants and the peak.

For the question given above, Activation energy = 20 + 50

Activation energy = +70 KJmol¯¹

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

a) the particles of the chair are strongly held together.