1(a) when 25.0ml of 0.500 mol.dm3 H2SO4(AQ) SOLUTION IS added to 0.025dm3 of 1.00 moldm3 KOH SOLUTION , THE TEMPERATURE OF THE REACTION MIXTURE RISES FROM 23.50 CELCIUS TO 30.17 CELCIUS . THE SPECIFIC HEAT CAPACITY AND DENSITY OF WATER 4.20j.g.celcius and 1.00 g.cm. the equation of the reaction is.
H2SO4 aq +2KOH aq =K2SO4 (aq) +2H2O
(I) IS THE REACTION ENDTHERMIC OR EXOTHERMIC? WHY?
(II) CALCULATE THE AMOUNT OF REACTANTS AVAILABLE (NUMBER OF MOLES) BEFORE THE REACTION . ARE THESE AMOUNTS IN A STOICHIOMETRIC(3sig. fig?
(iii) calculate the volume of water produced by the reaction and state whether it is significant or not if the overall volume of the reaction mixture is calculate with the correct number of decimal place . what is the total volume of the reaction mixture after the reaction is completed(3 sig.fig)?
(iv) calculate the amount of heat that was tranferred to the water(3 sig.fig)?
(V) calculate DH(DELTA H) THE ENTHALPY OF REACTION OF THIS NEUTRALISATION ( 3 SIG . FIG?
(VI) WHAT INCREASE OF TEMPERATURE WOULD BE OBSERVED IF 0.5 MOL OF SULPHURIC ACID REACTED WITH 1 MOL OF AQUEOUS POTASSIUM HYDROXIDE AND THE FINAL VOLUME OF THE REACTION MIXTURE WAS 1.0dm3?
Do you wish to just check your answers or do you not understand something about the problem? Please don't answer in caps; it makes it difficult to read.
(I) To determine if the reaction is endothermic or exothermic, we need to analyze the change in temperature. In this case, the temperature of the reaction mixture increased from 23.50°C to 30.17°C. An increase in temperature indicates that heat is being absorbed by the system, implying that the reaction is endothermic.
(II) To calculate the amount of reactants available before the reaction, we need to use the equation and the given concentrations. The balanced equation is:
H2SO4(aq) + 2KOH(aq) -> K2SO4(aq) + 2H2O
From the equation, we can see that the mole ratio between H2SO4 and KOH is 1:2. Using the given volume and molarity:
Volume of H2SO4(aq) = 25.0 mL = 0.0250 dm3
Concentration of H2SO4(aq) = 0.500 mol/dm3
Therefore, the number of moles of H2SO4 can be calculated as:
Moles of H2SO4 = Volume of H2SO4(aq) * Concentration of H2SO4(aq) = 0.0250 dm3 * 0.500 mol/dm3
Similarly, for KOH:
Volume of KOH(aq) = 0.025 dm3
Concentration of KOH(aq) = 1.00 mol/dm3
Moles of KOH = Volume of KOH(aq) * Concentration of KOH(aq) = 0.025 dm3 * 1.00 mol/dm3
These values will provide the number of moles for each reactant, and you can determine if the amounts are in stoichiometric proportions. Comparing the mole ratios in the balanced equation to the calculated moles will give you the answer.
(iii) To calculate the volume of water produced, we need to consider the stoichiometry of the reaction. From the balanced equation, we see that for every mole of H2SO4 reacted, 2 moles of H2O are formed. Therefore, the moles of water produced would be double the moles of H2SO4 reacted. We've already calculated the moles of H2SO4 in part (II), so the moles of water produced will be:
Moles of water = 2 * Moles of H2SO4
To calculate the volume, we can use the density of water. Given that the density of water is 1.00 g/cm3:
Volume of water = Moles of water * Molar mass of water / Density of water
(iv) To calculate the amount of heat transferred to the water, we can use the equation:
q = m * c * ΔT
where q is the heat transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. From the given information, we know the specific heat capacity of water is 4.20 J/g°C. We need to determine the mass of water using the density and volume of water produced in part (iii).
Mass of water = Volume of water * Density of water
Then the heat transferred to the water can be calculated using the formula.
(v) To calculate ΔH (enthalpy change or heat of reaction), we need to know the amount of heat transferred (calculated in part (iv)) and the moles of the limiting reactant (which can be obtained from part (II)). The relation between the amount of heat transferred and ΔH is given by:
ΔH = q / moles of limiting reactant
(vi) To calculate the increase in temperature in this scenario, where the mole ratio of H2SO4 to KOH is 1:1, we can use the concept of heat transfer being equal to the product of the number of moles and the change in enthalpy. Given that the mole ratio is 1:1, the heat transferred will be the same as in part (iv). Then, we can use the equation:
q = moles * ΔH
Substituting the known values (0.5 mol for moles and the ΔH calculated in part (v)), we can solve for the change in temperature (ΔT).
These steps should help you answer each part of the question.