Select the correct statement about thermochemical equations from the following: Select one: a. The specific heat capacity of all substances involved in the chemical reaction are listed to the right of the balanced chemical equation.
b. Because the enthalpy change of the reaction depends on the states of matter of the substances, they must be shown for each substance in the equation.
c. Positive values for the enthalpy change of a reaction indicate that the reaction released energy.
d. If the chemical reaction is reversed, then the arithmetic sign of the change in enthalpy stays the same.

The correct statement about thermochemical equations is  the enthalpy

change of the reaction depends on the states of matter of the substances,

they must be shown for each substance in the equation.

Enthalpy is defined as the heat change that occurs in a reaction or the sum

of the internal energy present. It is associated with change in matter and  is

dependent on the physical states of the reactants and products.

This is the reason why the physical states such as solid, liquid, gas etc have

to be shown in a thermochemical equation.

Read more about Enthalpy here link

Because the enthalpy change of the reaction depends on the states of matter of the substances, they must be shown for each substance in the equation.

Explanation:

The enthalpy of a substance depends on the particular state of matter in which it occurs.

Therefore, in a thermo chemical equation, all the states of matters for all the chemical species in the equation must be accurately written so that the thermo chemical equation can be meaningful.

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